Unit 19 · Algebra I: Exponents, Polynomials and Quadratics

A power is a short way to write repeated multiplication. 2^5 means 2 × 2 × 2 × 2 × 2, which equals 32. The 2 is the base, the number being multiplied, and the 5 is the exponent, the count of factors. Suppose a computer file doubles in size 3 times, then doubles 4 more times. The size grows by 2^3 and then by 2^4. Written out, that is (2 × 2 × 2) × (2 × 2 × 2 × 2), which is seven 2s in a row, or 2^7 = 128.

That example shows the product rule: when you multiply powers with the same base, keep the base and add the exponents. 2^3 × 2^4 = 2^(3 + 4) = 2^7. Check it with numbers: 8 × 16 = 128, and 2^7 = 128. The rule works for any base, including variables: x^2 × x^5 = x^7, and 3a^4 × 5a^2 = 15a^6. Multiply the coefficients (3 × 5 = 15), then add the exponents on a (4 + 2 = 6).

Division goes the other way. 5^6 ÷ 5^2 means six 5s on top and two 5s on the bottom. Two pairs cancel, leaving four 5s: 5^4 = 625. This is the quotient rule: keep the base and subtract the exponents. 5^6 ÷ 5^2 = 5^(6 − 2) = 5^4. Check: 15,625 ÷ 25 = 625. With variables, y^9 ÷ y^4 = y^5.

The most common mistake is multiplying the exponents instead of adding them. 2^3 × 2^4 is not 2^12; that would be 4,096, not 128. Another mistake is adding exponents when the bases differ. 2^3 × 5^2 cannot be combined into one power. Just compute it: 8 × 25 = 200. The rules apply only when the bases match.

Picture a square poster whose side is 2^3 = 8 inches long. Its area is (2^3)^2 = 8 × 8 = 64 square inches. Now count the 2s: 8 × 8 is (2 × 2 × 2) × (2 × 2 × 2), six 2s, so the area is 2^6 = 64. This is the power rule: to raise a power to a power, multiply the exponents. (2^3)^2 = 2^(3 × 2) = 2^6. Likewise (x^4)^5 = x^20.

The rule extends to products inside parentheses. (3x^2)^3 means 3x^2 × 3x^2 × 3x^2. Cube the 3 to get 27, and multiply the exponents on x: x^(2 × 3) = x^6. So (3x^2)^3 = 27x^6. A common mistake is writing 3x^6, forgetting to cube the coefficient. Every factor inside the parentheses gets the outside exponent.

What does 5^0 mean? Use the quotient rule: 5^3 ÷ 5^3 = 5^(3 − 3) = 5^0. But 5^3 ÷ 5^3 = 125 ÷ 125 = 1. So 5^0 = 1. Any nonzero base raised to the zero exponent is 1. Another way to see it: follow the pattern 5^3 = 125, 5^2 = 25, 5^1 = 5. Each step down divides by 5, so the next step is 5 ÷ 5 = 1.

Students often guess 5^0 = 0 because zero means nothing. But the exponent counts factors, and using zero factors of 5 leaves the starting value, 1, the number that leaves multiplication unchanged. Watch the parentheses: 3x^0 = 3 × 1 = 3, but (3x)^0 = 1, because the parentheses put the 3 inside the power.

Continue the pattern from 2^3 = 8, 2^2 = 4, 2^1 = 2, 2^0 = 1. Each step divides by 2. The next step is 1 ÷ 2 = 1/2, then 1/4, then 1/8. So 2^(−1) = 1/2, 2^(−2) = 1/4, 2^(−3) = 1/8. A negative exponent does not make a number negative. It tells you to take the reciprocal: flip the power into a fraction. 2^(−3) = 1/2^3 = 1/8.

The quotient rule agrees. 4^2 ÷ 4^5 = 4^(2 − 5) = 4^(−3). Written out, 16 ÷ 1,024 = 1/64, and 1/4^3 = 1/64. The same idea works with variables: x^3 ÷ x^7 = x^(−4) = 1/x^4. To write an expression with only positive exponents, move each factor with a negative exponent across the fraction bar and make its exponent positive. So 5x^(−2) = 5/x^2, and 1/y^(−3) = y^3.

Scientists use negative powers of ten to write tiny numbers. A red blood cell is about 7 × 10^(−6) meters across. Since 10^(−6) = 1/1,000,000, that is 0.000007 meters. A common mistake is thinking 10^(−6) is a negative number. It is small, not negative: a positive number less than 1.

Two errors to watch. First, (−3)^2 = 9 but 3^(−2) = 1/9; the position of the minus sign changes everything. Second, only the factor with the negative exponent flips. In 3x^(−2), the 3 stays on top: 3/x^2, not 1/(3x^2). The exponent belongs to x alone.

A square root of a number is a value that, times itself, gives the number. √49 = 7 because 7 × 7 = 49. A cube root uses the value three times: the cube root of 27 is 3 because 3 × 3 × 3 = 27. Roots undo powers. Now ask what 9^(1/2) could mean. By the product rule, 9^(1/2) × 9^(1/2) = 9^(1/2 + 1/2) = 9^1 = 9. So 9^(1/2) is a number that, squared, gives 9. That is the square root: 9^(1/2) = √9 = 3.

In general, b^(1/n) is the nth root of b, and b^(m/n) means take the nth root, then raise to the mth power, or the other way around; the order does not matter. Example: 8^(2/3). The cube root of 8 is 2, and 2^2 = 4, so 8^(2/3) = 4. Check by the other order: 8^2 = 64, and the cube root of 64 is 4. Same answer. Another: 16^(3/4). The fourth root of 16 is 2, since 2^4 = 16, and 2^3 = 8.

A rational exponent follows all the exponent rules. 4^(1/2) × 4^(3/2) = 4^(4/2) = 4^2 = 16. Check: 2 × 8 = 16. The rules also simplify a radical. √72 = √(36 × 2) = √36 × √2 = 6√2, because 36 is the largest perfect square that divides 72. Taking the root of each factor separately is the power rule applied to a product.

The most common mistake is reading 8^(1/3) as 8 ÷ 3 or as 8 × 1/3. It is the cube root: 2. Another mistake is 25^(1/2) = 12.5. Halving is not the same as taking a square root; 25^(1/2) = 5. Always ask: what number, multiplied by itself n times, gives the base?

The penny from the chapter story doubles each day. Day 1: $0.01. Day 2: $0.02. Day 3: $0.04. Day 4: $0.08. Each day you multiply by 2, not add 2. After d doublings the amount is 0.01 × 2^d. After 10 doublings, 0.01 × 2^10 = 0.01 × 1,024 = $10.24. After 20, 0.01 × 1,048,576 = $10,485.76. This is exponential growth: the same multiplier used again and again.

The general form is y = a × b^x. Here a is the starting value, the value when x = 0; b is the growth factor, the number you multiply by each step; and x counts the steps. A bacteria colony that starts at 500 cells and triples every hour has y = 500 × 3^x. After 4 hours: 500 × 3^4 = 500 × 81 = 40,500 cells. Check by stepping: 500, 1,500, 4,500, 13,500, 40,500.

Compare with a linear pattern. If the colony gained 500 cells per hour instead, after 4 hours it would have 500 + 4 × 500 = 2,500 cells. Adding the same amount each step is linear; multiplying by the same factor each step is exponential. In a table, linear values have a constant difference and exponential values have a constant ratio. The ratio of neighbors in 500, 1,500, 4,500 is 1,500 ÷ 500 = 3 and 4,500 ÷ 1,500 = 3.

A common mistake is computing 500 × 3^4 as (500 × 3)^4 = 1,500^4. The exponent applies only to the growth factor, so evaluate 3^4 first, then multiply by 500. Another mistake is swapping the roles of the two numbers in 2^10: the 2 is what you multiply by, and the 10 is how many times.

Suppose a savings account holds $1,000 and earns 5% interest each year, added to the balance. After one year the interest is 0.05 × 1,000 = $50, so the balance is $1,050. Notice that 1,000 + 0.05 × 1,000 = 1,000 × (1 + 0.05) = 1,000 × 1.05. Growing by 5% is the same as multiplying by 1.05. In the second year the 5% is figured on $1,050, so the interest is $52.50 and the balance is $1,102.50. Interest that earns interest is called compound interest.

The pattern is y = 1,000 × 1.05^t, where t is the number of years and 1,000 is the principal. After 10 years, 1.05^10 ≈ 1.629, so the balance is about $1,629. Simple interest, which pays $50 every year with no compounding, would give 1,000 + 10 × 50 = $1,500. The gap widens over time. The rule for percent growth: the growth factor is 1 + r, where r is the rate written as a decimal. A 3% rate gives 1.03, a 12% rate gives 1.12, and a 150% rate gives 2.5.

Some accounts compound more than once a year. The formula A = P(1 + r/n)^(nt) splits the yearly rate r among n periods per year. At 6% compounded monthly, each month multiplies by 1 + 0.06/12 = 1.005, and two years hold 24 periods. For $500: 500 × 1.005^24 ≈ 500 × 1.1272 ≈ $563.58. Yearly compounding at 6% would give 500 × 1.06^2 = 500 × 1.1236 = $561.80, a little less.

The most common mistake is using 0.05 as the growth factor. Multiplying by 0.05 shrinks $1,000 to $50; you must multiply by 1.05 to keep the original and add 5%. A second mistake is writing a 5% rate as 1.5. Move the decimal two places: 5% = 0.05, so the factor is 1.05, not 1.5.

A new car bought for $30,000 loses about 15% of its value each year. Losing 15% leaves 85%, so each year multiplies the value by 0.85. After 1 year: 30,000 × 0.85 = $25,500. After 2 years: 25,500 × 0.85 = $21,675. The formula is y = 30,000 × 0.85^t. This is exponential decay: the same fraction disappears each period. The decay factor is 1 − r, here 1 − 0.15 = 0.85. After 5 years, 0.85^5 ≈ 0.4437, so the car is worth about $13,311.

Radioactive materials decay with a half-life: the time it takes for half of the material to be gone. Iodine-131, used in some medical treatments, has a half-life of about 8 days. Start with 80 milligrams. After 8 days, 40 mg remain. After 16 days, 20 mg. After 24 days, 10 mg. The formula is y = 80 × (1/2)^(t/8), where t is in days. The exponent t/8 counts how many half-lives have passed. At t = 24, the exponent is 3 and (1/2)^3 = 1/8, so 80 × 1/8 = 10 mg.

Doctors use the same idea for medicine in the body. If a 200 mg dose of a drug has a half-life of 6 hours, then 100 mg remain after 6 hours, 50 mg after 12 hours, and 25 mg after 18 hours. To find the amount after 9 hours, the exponent is 9/6 = 1.5, and (1/2)^1.5 ≈ 0.354, so about 71 mg remain. Halving repeatedly never reaches zero; the amount just gets closer and closer to it. On a graph, the curve flattens toward the horizontal axis without touching it.

A common mistake is multiplying by 0.15 for a 15% loss. That computes the amount lost, not the amount left. Subtract from 1 first. Another mistake in half-life problems is using t itself as the exponent instead of t divided by the half-life. Twenty-four days with an 8-day half-life is 3 halvings, not 24.

A gym runs a referral plan: 3 people join the first week, and each week the number of new members is 3 times the week before. Week 1: 3. Week 2: 9. Week 3: 27. Week 4: 81. The list 3, 9, 27, 81, … is a geometric sequence. Each term comes from multiplying the term before it by the same number, the common ratio, here 3. To find the ratio of any geometric sequence, divide any term by the one before it: 27 ÷ 9 = 3.

You can describe the sequence two ways. The recursive rule says how to get the next term from the current one: a(n) = 3 × a(n − 1), with a(1) = 3. The explicit rule lets you jump straight to any term: a(n) = 3 × 3^(n − 1). Term 6 is 3 × 3^5 = 3 × 243 = 729. Check by stepping from 81: 243, then 729. In general, a(n) = a(1) × r^(n − 1). The exponent is n − 1 because the first term has had zero multiplications.

Ratios can be fractions. The sequence 400, 100, 25, 6.25 has ratio 100 ÷ 400 = 1/4, and its explicit rule is a(n) = 400 × (1/4)^(n − 1). Ratios can be negative: 2, −6, 18, −54 has ratio −3, so the signs alternate. Compare an arithmetic sequence such as 5, 8, 11, 14, which adds 3 each time. Arithmetic sequences are linear patterns; geometric sequences are exponential patterns.

The most common mistake is using r^n instead of r^(n − 1). For 3, 9, 27, …, the formula 3 × 3^n gives 9 for the first term, which is wrong. Test your formula on n = 1 before trusting it. Another mistake is finding the ratio by subtracting: 9 − 3 = 6 is a difference, not a ratio.

A biology class counts yeast cells under a microscope every hour and records: hour 0, 40 cells; hour 1, 60; hour 2, 90; hour 3, 135. Is the growth exponential? Check the ratios: 60 ÷ 40 = 1.5, 90 ÷ 60 = 1.5, 135 ÷ 90 = 1.5. The ratio is constant, so yes. The starting value is 40 and the growth factor is 1.5, so the model is y = 40 × 1.5^x. Predict hour 6: 40 × 1.5^6 = 40 × 11.390625 ≈ 456 cells.

Sometimes the table skips x = 0. A table shows x = 2, y = 50 and x = 3, y = 100. The ratio is 2. Work backward: at x = 1, y = 25; at x = 0, y = 12.5. So y = 12.5 × 2^x. Check: 12.5 × 2^2 = 50. If the steps in x are bigger than 1, divide carefully. Values of 10 at x = 0 and 90 at x = 2 mean the factor for two steps is 9, so the one-step factor is √9 = 3.

Stories work the same way. A video has 500 views, and views grow 20% per day. The starting value is 500, the growth factor is 1 + 0.20 = 1.2, and the model is y = 500 × 1.2^x. After 3 days: 500 × 1.2^3 = 500 × 1.728 = 864 views. To interpret a given model, read its parts. In y = 250 × 0.9^x, the 250 is the start, and the 0.9 means a 10% loss each step, because 1 − 0.9 = 0.10.

A common mistake is using the first value in the table as the starting value even when the table begins at x = 1. Another is checking differences instead of ratios. The yeast table has differences 20, 30, 45, which are not constant, so a line would not fit. Whenever values seem to speed up, test for a constant ratio.

Two job offers. Job A pays $100 the first day and adds $50 each day. Job B pays $1 the first day and doubles each day. On day 5, Job A pays 100 + 4 × 50 = $300 and Job B pays 1 × 2^4 = $16. Job A looks far better. But on day 10, Job A pays 100 + 9 × 50 = $550 and Job B pays 2^9 = $512. On day 12, Job A pays $650 and Job B pays 2^11 = $2,048. From then on, Job B wins by more every day.

This always happens. A linear function y = mx + b adds the same amount m each step; m is the slope. An exponential function y = a × b^x with b > 1 multiplies by b each step, so its increases themselves grow. Doubling from 512 adds 512; doubling from 2,048 adds 2,048. Eventually those jumps are larger than any fixed step, so the exponential function passes the linear one and stays ahead. On a graph, the line rises at a steady slope while the curve bends upward more and more steeply.

To decide which type fits data, compare steps of equal size. The list 5, 8, 11, 14 has a constant difference of 3, so it is linear: y = 3x + 5. The list 5, 10, 20, 40 has a constant ratio of 2, so it is exponential: y = 5 × 2^x. The list 5, 8, 13, 20 has neither: the differences 3, 5, 7 grow, and the ratios 1.6, 1.625, 1.54 are not constant. That list is quadratic, a shape you will meet in the last chapter of this unit.

A common mistake is judging by the first few values. Early on, a linear function with a big starting value or a steep slope can be far ahead, as Job A was for 10 days. Look at the long run. Another mistake is calling a pattern exponential because it grows fast. Fast is not the test; a constant ratio is.

A rectangular garden with side lengths x + 4 and x has area x(x + 4) = x^2 + 4x. That expression is a polynomial: a sum of terms, where each term is a number times a variable raised to a whole-number power. The number in front, like the 4 in 4x, is the coefficient. A term with no variable, like 7, is a constant term. Expressions with a variable under a root or in a denominator, like √x or 3/x, are not polynomials.

Polynomials get names from their number of terms and their degree. The degree of a term is its exponent, and the degree of the polynomial is the largest degree among its terms. 5x^3 − 2x + 1 has three terms, so it is a trinomial, and its degree is 3, which is called cubic. 4x^2 is one term, a monomial, of degree 2, called quadratic. x + 6 is a binomial of degree 1, called linear. A constant like 9 has degree 0.

Standard form writes the terms from highest degree to lowest. 3 + 2x^2 − 7x becomes 2x^2 − 7x + 3. The first coefficient, 2, is the leading coefficient. Before writing standard form, combine like terms: 4x^2 + x − x^2 + 5x = 3x^2 + 6x. Like terms have the same variable with the same exponent. The terms 3x^2 and 6x are not like terms, so they stay separate.

A common mistake is reading the degree from the first term as written. In 5x + 2x^4 − 1 the degree is 4, not 1, because degree depends on the largest exponent anywhere. Another is combining 3x^2 + 6x into 9x^3, which mixes unlike terms. Adding coefficients works only when the variable parts match exactly.

A school store sells notebooks and pens. Monday’s income is 3n + 2p dollars, where n is the price of a notebook and p is the price of a pen. Tuesday’s income is 5n + 4p. The total is (3n + 2p) + (5n + 4p) = 8n + 6p. Adding polynomials means adding the like terms: 3n + 5n = 8n and 2p + 4p = 6p. With powers it works the same way: (2x^2 + 3x − 1) + (x^2 − 5x + 4) = 3x^2 − 2x + 3. Line up the like terms: 2x^2 + x^2 = 3x^2, 3x − 5x = −2x, and −1 + 4 = 3.

Subtraction needs one extra step. (4x^2 + x − 6) − (x^2 − 3x + 2) asks you to take away the whole second polynomial, every term of it. Distribute the minus sign: −(x^2 − 3x + 2) = −x^2 + 3x − 2. Now add: 4x^2 − x^2 = 3x^2, x + 3x = 4x, and −6 − 2 = −8. The result is 3x^2 + 4x − 8. Check with x = 1: the original is (4 + 1 − 6) − (1 − 3 + 2) = −1 − 0 = −1, and 3 + 4 − 8 = −1. It matches.

Substituting a number is a good check on any polynomial arithmetic. Degree is another check: adding two quadratics never gives a degree above 2, though it can give less, since the x^2 terms can cancel: (x^2 + 5) + (−x^2 + x) = x + 5. Polynomials are closed under addition and subtraction, which means the result is always another polynomial, just as adding two integers always gives an integer.

The classic mistake is subtracting only the first term: writing 4x^2 − x^2 − 3x + 2 and getting 3x^2 − 3x + 2. The minus sign must reach the −3x and the +2 and flip both of them. Write the sign change on its own line before combining anything, and then substitute a number to make sure the result matches.

A rectangle is x + 3 units long and x + 2 units wide. Its area is (x + 3)(x + 2). Split the rectangle into four pieces: an x by x square, an x by 2 strip, a 3 by x strip, and a 3 by 2 corner. The pieces have areas x^2, 2x, 3x and 6, so the total is x^2 + 5x + 6. This area model shows why multiplying two binomials produces four products before combining. Each term of the first factor meets each term of the second.

Without the picture, use the distributive property twice: (x + 3)(x + 2) = x(x + 2) + 3(x + 2) = x^2 + 2x + 3x + 6 = x^2 + 5x + 6. Some students remember the four products as First, Outer, Inner, Last. Check with x = 10: 13 × 12 = 156, and 100 + 50 + 6 = 156. Multiplying by a monomial is simpler: 3x(2x^2 − 5x + 1) = 6x^3 − 15x^2 + 3x. Multiply the coefficients and add the exponents in each term.

Signs need care. (2x − 5)(x + 4) = 2x^2 + 8x − 5x − 20 = 2x^2 + 3x − 20. The middle terms have opposite signs, so they partly cancel. (x − 3)(x − 6) = x^2 − 6x − 3x + 18 = x^2 − 9x + 18; two negatives make the last term positive. The area model still works with negatives. Treat each box as a signed product.

The most common mistake is multiplying only the first terms and the last terms: (x + 3)(x + 2) = x^2 + 6. That skips the two strips, the outer and inner products, which together give 5x. Draw the four boxes until the pattern is automatic. A second mistake is adding coefficients when they should be multiplied: 2x × 4x = 8x^2, not 6x^2.

Square a binomial with the area model: (x + 5)^2 = (x + 5)(x + 5). The four boxes are x^2, 5x, 5x and 25, so (x + 5)^2 = x^2 + 10x + 25. The middle term is twice the product of the two parts. In general, (a + b)^2 = a^2 + 2ab + b^2 and (a − b)^2 = a^2 − 2ab + b^2. The result is a perfect square trinomial. Example: (3x − 4)^2 = 9x^2 − 24x + 16, because 2 × 3x × 4 = 24x. Check with x = 2: (6 − 4)^2 = 4, and 36 − 48 + 16 = 4.

Now multiply a sum by a difference: (x + 6)(x − 6) = x^2 − 6x + 6x − 36 = x^2 − 36. The middle terms cancel every time. This is the difference of squares pattern: (a + b)(a − b) = a^2 − b^2. It gives a mental-math trick: 23 × 17 = (20 + 3)(20 − 3) = 400 − 9 = 391. With variables, (2x + 7)(2x − 7) = 4x^2 − 49.

These patterns are examples of an identity: an equation true for every value of the variable, not just for one solution. You can test an identity by substituting several numbers, but proving it means expanding both sides and showing they match term for term. Identities reveal structure. Seeing x^2 − 36 as x squared minus 6 squared, or 4x^2 + 12x + 9 as (2x + 3) squared, is the key to factoring in the next section.

The most common mistake in all of algebra may be (x + 5)^2 = x^2 + 25. That drops the middle term 10x. Test it: x = 1 gives 36 on the left and 26 on the right, so the shortcut is false. Squaring a sum is not the same as adding the squares. Draw the four boxes if you are ever unsure.

Multiplying 3x(2x + 5) gives 6x^2 + 15x. Factoring runs that backward: start from 6x^2 + 15x and write it as a product. Ask what every term shares. The coefficients 6 and 15 share 3. The variable parts x^2 and x share x. So the greatest common factor, or GCF, is 3x. Divide each term by it: 6x^2 ÷ 3x = 2x and 15x ÷ 3x = 5. Write 6x^2 + 15x = 3x(2x + 5). Multiply back out to check.

Try 12x^3 − 18x^2 + 6x. The GCF of 12, 18 and 6 is 6, and the lowest power of x present is x, so the GCF is 6x. Dividing gives 2x^2 − 3x + 1, so the factored form is 6x(2x^2 − 3x + 1). Notice the 1 at the end: 6x ÷ 6x = 1, and it must stay. Without it the factored form would multiply back to 12x^3 − 18x^2, with the last term missing. That lost 1 is the most common GCF mistake.

If the leading coefficient is negative, factor out a negative GCF so the first term inside is positive: −4x^2 + 8x = −4x(x − 2). Sometimes the common factor is a whole binomial. In 5(x + 3) + x(x + 3), the two terms share (x + 3), so the expression equals (x + 3)(5 + x). That move powers factoring by grouping later in this chapter.

A second mistake is pulling out a factor that is common but not the greatest: 6x^2 + 15x = 3(2x^2 + 5x) is true but unfinished, since 2x^2 + 5x still shares an x. To factor completely, check that the terms in parentheses have no common factor left and that no factor can be broken down further.

A rectangle has area x^2 + 7x + 12. What are its side lengths? Since (x + p)(x + q) = x^2 + (p + q)x + pq, you need two numbers that multiply to 12 and add to 7. List each factor pair of 12: 1 and 12 (sum 13), 2 and 6 (sum 8), 3 and 4 (sum 7). So x^2 + 7x + 12 = (x + 3)(x + 4). Check by multiplying: x^2 + 4x + 3x + 12. It matches.

Signs tell you where to look. In x^2 − 8x + 15, the product 15 is positive and the sum −8 is negative, so both numbers are negative: −3 and −5. The factors are (x − 3)(x − 5). In x^2 + 2x − 15, the product is negative, so one number is positive and one is negative; they are 5 and −3, since 5 + (−3) = 2. The factors are (x + 5)(x − 3). In x^2 − 2x − 15 the signs swap: (x − 5)(x + 3).

Not every trinomial factors with integers. x^2 + 3x + 5 needs two numbers with product 5 and sum 3, but the only pairs are 1 and 5 (sum 6) or −1 and −5 (sum −6). So it is a prime polynomial over the integers. Later you will still solve equations like x^2 + 3x + 5 = 0 with other tools. Always factor out a GCF first: 2x^2 + 14x + 24 = 2(x^2 + 7x + 12) = 2(x + 3)(x + 4).

A common mistake is finding numbers that multiply to c but forgetting the sum: (x + 2)(x + 6) multiplies to x^2 + 8x + 12, not x^2 + 7x + 12. Always multiply your answer back out. Another mistake is losing a sign: x^2 − 8x + 15 is (x − 3)(x − 5), not (x + 3)(x + 5).

Factor 2x^2 + 7x + 3. The leading coefficient is 2, so guessing (2x + ?)(x + ?) is possible but slow. A reliable method: multiply a × c = 2 × 3 = 6, then find two numbers with product 6 and sum 7. They are 1 and 6. Split the middle term: 2x^2 + x + 6x + 3. Group in pairs: (2x^2 + x) + (6x + 3) = x(2x + 1) + 3(2x + 1). Both groups share (2x + 1), so the answer is (2x + 1)(x + 3). Multiply back: 2x^2 + 6x + x + 3. It matches.

Try 6x^2 − 11x + 4. Here a × c = 24 and the sum is −11, so both numbers are negative: −3 and −8 (product 24, sum −11). Split: 6x^2 − 3x − 8x + 4. Group: 3x(2x − 1) − 4(2x − 1). Note that −4 was factored out of the second pair, so that −8x + 4 = −4(2x − 1). Answer: (2x − 1)(3x − 4). Check the outer and inner products: 2x × (−4) + (−1)(3x) = −8x − 3x = −11x.

The order of the split does not matter. Splitting 7x as 6x + x gives (2x^2 + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1), the same result. If the grouping fails to produce a shared binomial, either the two numbers are wrong or the trinomial is prime. And as always, pull out a GCF first: 4x^2 + 14x + 6 = 2(2x^2 + 7x + 3) = 2(2x + 1)(x + 3).

The most common mistake is using c instead of a × c when hunting for the pair. For 2x^2 + 7x + 3, looking for a product of 3 gives 1 and 3, which sum to 4, not 7. Another is dropping a negative when factoring the second pair: −8x + 4 = −4(2x − 1), not 4(2x − 1).

x^2 − 49 has no middle term. Its two terms are perfect squares, x^2 and 7^2, separated by a minus sign, so it is a difference of squares: x^2 − 49 = (x + 7)(x − 7). Check: the middle terms −7x and +7x cancel. The pattern works with coefficients too: 9x^2 − 25 = (3x)^2 − 5^2 = (3x + 5)(3x − 5). And 50x^2 − 8 = 2(25x^2 − 4) = 2(5x + 2)(5x − 2), after pulling out the GCF.

A sum of squares like x^2 + 49 does not factor with real numbers. Test it: (x + 7)(x + 7) = x^2 + 14x + 49 and (x + 7)(x − 7) = x^2 − 49; neither works. Leave x^2 + 49 alone, and do not confuse it with x^2 − 49. The minus sign is what makes the pattern work.

A perfect square trinomial is the other pattern. x^2 + 12x + 36 has a perfect square at each end, x^2 and 6^2, and the middle term is 2 × 6 × x = 12x. So it equals (x + 6)^2. For 4x^2 − 20x + 25, the ends are (2x)^2 and 5^2, and 2 × 2x × 5 = 20x matches the middle, so it is (2x − 5)^2, with the minus sign following the middle term. If the middle does not match, use the regular trinomial method: x^2 + 10x + 36 is not a perfect square, since 2 × 6 = 12, not 10.

Recognizing these patterns saves time and reveals structure. A common mistake is writing x^2 − 49 as (x − 7)^2; multiply back to see it gives x^2 − 14x + 49. Another is stopping early: x^4 − 16 = (x^2 + 4)(x^2 − 4), and x^2 − 4 factors again to (x + 2)(x − 2), so the complete answer is (x^2 + 4)(x + 2)(x − 2).

x^3 + 3x^2 + 4x + 12 has four terms and no factor common to all four. Group the terms in pairs: (x^3 + 3x^2) + (4x + 12). Factor each pair: x^2(x + 3) + 4(x + 3). Now both parts share (x + 3), a common binomial factor. Pull it out: (x + 3)(x^2 + 4). Multiply back: x^3 + 4x + 3x^2 + 12, the original polynomial with its terms in a different order.

Grouping needs the right pairing and the right signs. In 2x^3 − 6x^2 − 5x + 15, the pairs are (2x^3 − 6x^2) + (−5x + 15). The first pair gives 2x^2(x − 3). For the second pair, factor out −5 so the binomial matches: −5(x − 3). Then the expression is (x − 3)(2x^2 − 5). If you factored out +5 instead, you would get 5(−x + 3), which does not match (x − 3), and grouping would seem to fail.

Grouping is the engine behind the ax^2 + bx + c method from earlier: splitting the middle term turns a trinomial into four terms, and then you group. It also handles two variables: xy + 2x + 3y + 6 = x(y + 2) + 3(y + 2) = (y + 2)(x + 3). When the first pairing fails, try another order. For x^3 + 2x + 5x^2 + 10, reorder to x^3 + 5x^2 + 2x + 10 = x^2(x + 5) + 2(x + 5) = (x + 5)(x^2 + 2).

Here is a factoring plan for any polynomial. First, factor out the GCF. Second, count terms: two terms, check for a difference of squares; three terms, try a perfect square or the trinomial methods; four terms, group. Third, check whether any factor breaks down further. Fourth, multiply back to confirm. A common mistake is stopping at x^2(x + 3) + 4(x + 3). That is a sum, not a product, until you pull out (x + 3).

Back to the garden path. Its area is 4x^2 + 32x, and it must hold exactly 80 square feet of gravel: 4x^2 + 32x = 80. Move everything to one side: 4x^2 + 32x − 80 = 0. Factor: 4(x^2 + 8x − 20) = 4(x + 10)(x − 2) = 0. Now use the zero-product property: if two numbers multiply to zero, one of them must be zero. So x + 10 = 0 or x − 2 = 0, which gives x = −10 or x = 2. A width cannot be negative, so the path is 2 feet wide. Check: 4(4) + 32(2) = 16 + 64 = 80.

The property works only when one side is exactly 0. (x − 3)(x + 1) = 5 tells you nothing about the factors separately, because many pairs of numbers multiply to 5. Expand instead: x^2 − 2x − 3 = 5, then x^2 − 2x − 8 = 0, then (x − 4)(x + 2) = 0, so x = 4 or x = −2. Check x = 4: (1)(5) = 5. Check x = −2: (−5)(−1) = 5. Both work.

The solutions of a polynomial equation are also called a root of the equation or a zero of a function. They show where the graph crosses the x-axis, which is why factored form is so useful: f(x) = (x − 4)(x + 2) has zeros at 4 and −2 that you can read directly. A perfect square such as x^2 − 6x + 9 = (x − 3)^2 = 0 has one repeated root, x = 3. An equation like x^2 = 5x also yields to factoring: x^2 − 5x = 0, so x(x − 5) = 0 and x = 0 or x = 5.

In that last example, the common mistake is dividing both sides by x to get x = 5. Dividing by x throws away the solution x = 0. Never divide an equation by an expression that could be zero; factor it out instead. A second mistake is solving (x + 10)(x − 2) = 0 as x = 10 and x = −2. Each factor set to zero gives the opposite sign of its constant.

Take f(x) = x^2 − 4x + 3 and make a table: x = 0 gives 3, x = 1 gives 0, x = 2 gives −1, x = 3 gives 0, x = 4 gives 3, and x = 5 gives 8. Plot the points and connect them with a smooth curve. The U shape is a parabola. The lowest point, (2, −1), is the vertex. The vertical line x = 2 through the vertex is the axis of symmetry. Points the same distance on either side have equal heights: f(1) = f(3) = 0 and f(0) = f(4) = 3.

A quadratic function is any function of the form f(x) = ax^2 + bx + c with a not equal to 0. When a is positive, the parabola opens upward and the vertex is the minimum. When a is negative, it opens downward and the vertex is the maximum, as with a thrown ball. The size of a controls the width: y = 3x^2 is narrower than y = x^2, and y = 0.25x^2 is wider. Compare at x = 2: the outputs are 12, 4 and 1.

Intercepts anchor the graph. The y-intercept is f(0) = c, here 3. An x-intercept is where f(x) = 0, here x = 1 and x = 3; these are the zeros of the function. The axis of symmetry passes exactly halfway between the two x-intercepts: (1 + 3)/2 = 2. Not every parabola crosses the x-axis. y = x^2 + 1 has vertex (0, 1) and opens upward, so it never reaches y = 0 and has no x-intercepts.

Common mistakes: plotting too few points and drawing a V instead of a smooth curve, since the parabola is rounded at the vertex; and assuming the vertex always sits on the y-axis. Only parabolas whose equation has no x term, with b = 0, have their axis at x = 0.

A toy rocket launched from the ground has height h = −16t^2 + 48t feet after t seconds. When is it highest? The equation is in standard form y = ax^2 + bx + c with a = −16, b = 48, c = 0. The vertex formula gives the axis of symmetry: x = −b/(2a) = −48/(2 × −16) = −48/(−32) = 1.5. The rocket is highest at t = 1.5 seconds. Its height then is h = −16(1.5)^2 + 48(1.5) = −16(2.25) + 72 = −36 + 72 = 36 feet. The vertex is (1.5, 36), a maximum since a is negative.

Why does −b/(2a) work? The zeros of −16t^2 + 48t = −16t(t − 3) are t = 0 and t = 3, and the axis is halfway between them: 1.5. The formula gives the same answer without factoring, even when the zeros are hard to find. Example: y = 2x^2 + 8x + 5. Axis: x = −8/(2 × 2) = −2. Vertex y: 2(4) + 8(−2) + 5 = 8 − 16 + 5 = −3. The vertex is (−2, −3), a minimum since a = 2 is positive.

To sketch from standard form: find the vertex, mark the y-intercept (0, c), then reflect that point across the axis. For y = 2x^2 + 8x + 5, the y-intercept (0, 5) reflects across x = −2 to (−4, 5). Three points and the direction of opening give a good sketch. The maximum or minimum value of the function is the y-coordinate of the vertex; the x-coordinate is where it happens. Keep those straight: the rocket’s maximum height is 36 feet, reached at 1.5 seconds.

The most common error is a sign slip in −b/(2a). For y = x^2 − 6x + 1, b = −6, so x = −(−6)/(2 × 1) = 3, not −3. Another error is stopping at the x-coordinate and reporting it as the maximum value. Always substitute back to get y; the vertex is a point with two coordinates.

Start with the parent function y = x^2, whose vertex is (0, 0). Adding a number outside the square slides the graph up: y = x^2 + 3 has vertex (0, 3). Subtracting inside the square slides it right: y = (x − 4)^2 has vertex (4, 0), because the squared part is zero when x = 4. Together, y = (x − 4)^2 + 3 has vertex (4, 3). This is vertex form: y = a(x − h)^2 + k, with vertex (h, k). Watch the sign: (x − 4) means h = 4, and (x + 2) means h = −2.

The number a makes a stretch or a flip. y = 2(x − 1)^2 − 5 has vertex (1, −5), opens upward, and is narrower than y = x^2. y = −0.5(x + 3)^2 + 2 has vertex (−3, 2), opens downward, and is wider. To graph, plot the vertex, then step sideways and use a. From (1, −5), moving 1 unit left or right raises the curve by 2 × 1^2 = 2, to (0, −3) and (2, −3). Moving 2 units raises it by 2 × 4 = 8, to (−1, 3) and (3, 3).

Converting vertex form to standard form is just expanding: y = 2(x − 1)^2 − 5 = 2(x^2 − 2x + 1) − 5 = 2x^2 − 4x + 2 − 5 = 2x^2 − 4x − 3. Check the vertex with the formula: x = 4/(2 × 2) = 1, which matches. Going the other way, from standard to vertex form, takes completing the square, which comes later in this chapter. Each form shows something different: vertex form shows the vertex, and standard form shows the y-intercept.

The most common mistake is reading (x − 4)^2 as a shift of 4 to the left. Ask what x makes the parentheses zero; that answer, 4, is the vertex’s x. A second mistake is forgetting the a when plotting points: for y = 2(x − 1)^2 − 5, the point one unit from the vertex is 2 higher, not 1.

A rectangular pen along a river uses 40 feet of fence on three sides, with the river as the fourth side. If the two sides that meet the river are x feet, the side parallel to it is 40 − 2x, and the area is A = x(40 − 2x). This is factored form: A = −2x(x − 20), with zeros at x = 0 and x = 20, since a pen with no width or no length has zero area. The vertex is halfway between the zeros, at x = 10. The maximum area is A = 10(40 − 20) = 200 square feet.

In general, y = a(x − p)(x − q) has zeros p and q, because a product is zero when a factor is zero. y = 2(x − 3)(x + 1) has zeros 3 and −1. The axis is x = (3 + (−1))/2 = 1, and the vertex y is 2(1 − 3)(1 + 1) = 2(−2)(2) = −8. The vertex is (1, −8). The y-intercept comes from x = 0: 2(−3)(1) = −6. To expand into standard form: 2(x^2 − 2x − 3) = 2x^2 − 4x − 6, and the constant −6 matches the y-intercept.

Going from standard form to factored form is the factoring from the previous chapter. y = x^2 − 2x − 15 = (x − 5)(x + 3), so each zero of a function is easy to read: 5 and −3, where the graph crosses the x-axis. Three forms, three views of one parabola: standard form gives the y-intercept, vertex form gives the vertex, factored form gives the x-intercepts. Choose the form that answers the question you are asked.

Common mistakes: reading zeros with the wrong sign, so (x − 5)(x + 3) is said to have zeros −5 and 3; and worrying about the a in front. The zeros of −3(x − 1)(x − 7) are 1 and 7. The a affects the height and direction of the curve, not where it crosses the axis.

A square patio has an area of 144 square feet. What is its side length? x^2 = 144, so x = √144 = 12 feet. But the equation x^2 = 144 has two solutions, 12 and −12, because (−12)^2 is also 144. We write x = ±12 and use the situation to choose: a side length is 12. This is the square root method: if x^2 = k with k > 0, then x = √k or x = −√k. The symbol ± means plus or minus.

Isolate the squared part first. 3x^2 − 27 = 0: add 27 to get 3x^2 = 27, divide by 3 to get x^2 = 9, so x = ±3. If the number is not a perfect square, simplify the radical: 2x^2 = 40 gives x^2 = 20, so x = ±√20 = ±2√5, about ±4.47. The method also works when a binomial is squared. (x − 5)^2 = 49: take roots, x − 5 = ±7, so x = 5 + 7 = 12 or x = 5 − 7 = −2. Check: (12 − 5)^2 = 49 and (−2 − 5)^2 = 49.

Dropped objects use this method. Ignoring air resistance, an object dropped from a height of h0 feet is at height h = h0 − 16t^2 after t seconds. From a 100-foot bridge, when does a dropped stone hit the water? 0 = 100 − 16t^2, so 16t^2 = 100, t^2 = 6.25, and t = 2.5 seconds; the negative root would be a time before the drop, so discard it. Check: 16 × 6.25 = 100.

The classic mistake is dropping the negative root when both matter. x^2 = 25 has solutions 5 and −5; leaving out −5 loses half the answer. The opposite mistake is inventing solutions: x^2 = −25 has no real solution, because no real number squared is negative. And never take the square root before isolating: from x^2 + 9 = 25, the expression √(x^2 + 9) is not x + 3.

The square root method needs a perfect square on one side. Can we make one? Take x^2 + 6x = 7. In the area model, x^2 + 6x is an x by x square plus two strips of 3x, one along the right and one along the bottom. The picture is a bigger square missing its corner, a 3 by 3 piece of area 9. Adding 9 fills the corner, which is why the method is called completing the square: x^2 + 6x + 9 = (x + 3)^2, a perfect square trinomial. To keep the equation balanced, add 9 to both sides: (x + 3)^2 = 16.

Now finish with square roots: x + 3 = ±4, so x = 1 or x = −7. Check: 1 + 6 = 7, and 49 − 42 = 7. The number to add is always (b/2)^2: half the coefficient of x, then squared. For x^2 − 10x = 11: half of −10 is −5, and (−5)^2 = 25. So x^2 − 10x + 25 = 36, which is (x − 5)^2 = 36. Then x − 5 = ±6, so x = 11 or x = −1. Check: 121 − 110 = 11 and 1 + 10 = 11.

If the leading coefficient is not 1, divide it out first. 2x^2 + 8x − 10 = 0 becomes x^2 + 4x − 5 = 0, then x^2 + 4x = 5, add 4: (x + 2)^2 = 9, x = 1 or x = −5. The same steps convert a function to vertex form. For y = x^2 + 4x − 5, add and subtract 4 on the right: y = (x^2 + 4x + 4) − 4 − 5 = (x + 2)^2 − 9. The vertex is (−2, −9), matching the formula x = −4/2 = −2.

Two common mistakes: adding (b/2)^2 to one side only, which changes the equation; and squaring b instead of b/2, so for x^2 + 6x someone adds 36 instead of 9. Say it aloud: half of six is three, three squared is nine. When b is odd, fractions appear: for x^2 + 3x, add (3/2)^2 = 9/4. That is fine; the method still works.

Completing the square on the general equation ax^2 + bx + c = 0, once and for all values, produces the quadratic formula: x = (−b ± √(b^2 − 4ac)) / (2a). It solves every quadratic, whether or not it factors. Example: 2x^2 + 3x − 5 = 0, so a = 2, b = 3, c = −5. First compute the part under the root: b^2 − 4ac = 9 − 4(2)(−5) = 9 + 40 = 49. Then x = (−3 ± 7)/4. Two answers: (−3 + 7)/4 = 1 and (−3 − 7)/4 = −10/4 = −2.5. Check x = 1: 2 + 3 − 5 = 0.

The formula needs the equation in the form ax^2 + bx + c = 0, with zero on one side. To solve x^2 = 6x − 4, first rewrite it: x^2 − 6x + 4 = 0, so a = 1, b = −6, c = 4. Under the root: 36 − 16 = 20. So x = (6 ± √20)/2. Leave it in radical form by simplifying √20 = 2√5, so x = (6 ± 2√5)/2 = 3 ± √5. As decimals, 3 + 2.236 = 5.236 and 3 − 2.236 = 0.764. Neither is a whole number, which is why factoring would have failed.

Which method should you use? If the equation has no x term, take square roots. If it factors easily, factor. If a = 1 and b is even, completing the square is quick. Otherwise, use the formula; it always works. In the chapter story, the shot is level with the rim when −16t^2 + 24t + 6 = 10, that is, −16t^2 + 24t − 4 = 0. Divide by −4: 4t^2 − 6t + 1 = 0. Then t = (6 ± √(36 − 16))/8 = (6 ± √20)/8, about 0.19 seconds and 1.31 seconds.

Common mistakes: forgetting that −b flips the sign of b (with b = −6, −b = 6); a sign error in −4ac when c is negative (−4 × 2 × −5 = +40); and dividing only the root by 2a instead of the whole numerator. Write the fraction bar under everything. Substituting the answers back into the original equation catches all three.

The part of the formula under the square root, b^2 − 4ac, is the discriminant. It decides how many real solutions there are before you solve. If it is positive, the ± gives two different real solutions. If it is zero, ± 0 gives one repeated root. If it is negative, there is no real square root, so there is no real solution. Example: x^2 − 6x + 9 = 0 has discriminant 36 − 36 = 0, one solution, x = 3, since it is (x − 3)^2. And x^2 + x + 1 = 0 has 1 − 4 = −3, no real solution.

On a graph, the discriminant counts x-intercepts. y = x^2 − 6x + 9 touches the x-axis once, at its vertex (3, 0). y = x^2 + x + 1 has its vertex at x = −1/2, where y = 1/4 − 1/2 + 1 = 3/4, above the axis and opening upward, so it never crosses. y = x^2 − 6x + 5 has discriminant 36 − 20 = 16, which is positive, and it crosses twice, at 1 and 5. A discriminant that is a perfect square, like 16, also signals that the quadratic factors with integers: (x − 1)(x − 5).

Real problems use this test. A ball thrown upward at 32 feet per second from 5 feet has height h = −16t^2 + 32t + 5. Does it ever reach 25 feet? Set h = 25: −16t^2 + 32t − 20 = 0. Discriminant: 32^2 − 4(−16)(−20) = 1,024 − 1,280 = −256, negative. So no, the ball never reaches 25 feet. Check with the vertex: t = 1, h = −16 + 32 + 5 = 21 feet at most. The discriminant answered without solving.

A common mistake is computing 4ac with the wrong sign. For a = −16 and c = −20, 4ac = +1,280, so b^2 − 4ac subtracts 1,280. Another is reporting one solution as no solution when the discriminant is zero. Zero is a fine square root; it just gives a single answer.

When an object is thrown upward, gravity pulls it back with the same acceleration no matter its size, ignoring air resistance. In feet and seconds, its height is h = −16t^2 + vt + h0, where v is the upward launch speed in feet per second and h0 is the initial height. In meters, the −16 becomes about −4.9. A basketball released 6 feet up at 24 feet per second follows h = −16t^2 + 24t + 6, the equation from the chapter story. Any such thrown object is a projectile.

Three questions, three tools. When is it highest? The vertex: t = −24/(2 × −16) = 0.75 seconds, and h = −16(0.5625) + 18 + 6 = −9 + 24 = 15 feet. When does it pass a given height, like the 10-foot rim? Set h = 10 and solve −16t^2 + 24t − 4 = 0 with the formula: about 0.19 and 1.31 seconds, on the way up and on the way down. When does it land, if it misses everything? Set h = 0: t = (−24 ± √(576 + 384))/(−32). Since √960 ≈ 30.98, t ≈ 1.72 seconds; the other root is negative.

Interpret each part of the equation. The 6 is the y-intercept, the height at t = 0. The 24 sets how steeply the ball rises at first. The −16 bends the path downward, so the parabola opens downward and the vertex is a maximum. A soccer ball kicked from the ground at 40 feet per second has h = −16t^2 + 40t. It is in the air until h = 0 again: −16t(t − 2.5) = 0, so 2.5 seconds. Its peak, at t = 1.25, is −16(1.5625) + 50 = −25 + 50 = 25 feet.

Common mistakes: using +16 instead of −16, which makes the ball rise forever; and reading the vertex’s t as the height. Also, this parabola is a graph of height against time, not a picture of the ball’s flight across the court, even though a thrown ball’s path through the air happens to be a parabola as well.

A poster is 3 inches longer than it is wide, and its area is 130 square inches. First define a variable: let the width be w. Then the length is w + 3 and the area is w(w + 3) = 130. Expand: w^2 + 3w − 130 = 0. Factor: two numbers with product −130 and sum 3 are 13 and −10, so (w + 13)(w − 10) = 0, and w = 10 or w = −13. A width cannot be negative, so reject a solution that fails the story: w = 10 inches and the length is 13 inches. Check: 10 × 13 = 130.

The garden border again: a 10 by 6 bed with a path of width x has total area (10 + 2x)(6 + 2x). If the whole thing must be 140 square feet: 60 + 20x + 12x + 4x^2 = 140, so 4x^2 + 32x − 80 = 0, then x^2 + 8x − 20 = 0, then (x + 10)(x − 2) = 0, so x = 2 feet. Check: 14 × 10 = 140. Number puzzles work the same way. The product of two consecutive integers is 72 means n(n + 1) = 72, so n^2 + n − 72 = 0, (n + 9)(n − 8) = 0, and the pairs are 8 and 9 or −9 and −8. Both pairs are valid here; nothing in the puzzle forbids negatives.

Revenue problems produce a maximum. A theater sells 200 tickets at $10. Each $1 raise in price loses 10 buyers. With x raises, the revenue is R = (10 + x)(200 − 10x) = 2,000 + 100x − 10x^2. The vertex is at x = −100/(2 × −10) = 5. So the price is $15, the buyers number 150, and the revenue is $2,250, better than the original $2,000. The parabola opens downward, so the vertex is the maximum revenue.

The steps: define a variable, write expressions for the other quantities, write an equation from the given total or relationship, solve, and check against the story. The most common mistake is keeping a negative or impossible solution. The second is stopping when the variable is found even though the question asked for something else, such as the length rather than the width.

Three tables, each starting at x = 0 with y = 1. Table A: 1, 3, 5, 7, 9. Table B: 1, 3, 9, 27, 81. Table C: 1, 3, 7, 13, 21. Table A has first differences 2, 2, 2, 2, so it is linear: y = 2x + 1. Table B has ratios 3, 3, 3, 3, so it is exponential: y = 3^x. Table C has first differences 2, 4, 6, 8, not constant, but its second differences, the differences of the differences, are 2, 2, 2. That is the quadratic pattern. Its rule is y = x^2 + x + 1; check x = 3: 9 + 3 + 1 = 13.

Why constant second differences? For y = ax^2 + bx + c with x stepping by 1, the second difference is always 2a. Table C has second difference 2, so a = 1. That is a fast way to find the leading coefficient. For the table 0, 3, 12, 27, 48 at x = 0 to 4, the first differences are 3, 9, 15, 21 and the second differences are 6, 6, 6, so a = 3 and the rule is y = 3x^2. Check x = 4: 3 × 16 = 48.

In the long run, the three types behave differently. Linear growth adds a fixed amount forever. Quadratic growth speeds up, but its increases grow only linearly. Exponential growth multiplies, so its increases grow exponentially, and it eventually outruns any quadratic, whatever the coefficients. Compare y = x^2 and y = 2^x: at x = 3, 9 versus 8; at x = 4, 16 versus 16; at x = 10, 100 versus 1,024. On a graph, the exponential curve crosses the parabola and leaves it behind.

A common mistake is calling anything curved exponential. Test with ratios: Table C has ratios 3, 2.33, 1.86, 1.62, not constant, so it is not exponential. Another mistake is stopping after the first differences. If they are not constant, take the second differences before deciding.