Chapter 14 — Dividing Polynomials and Equivalent Quadratic Forms

Standard: A.EO.2 (d, e)

A.EO.2 — verbatim. The student will perform operations on and factor polynomial expressions in one variable. Students will demonstrate the following Knowledge and Skills: a) Determine sums and differences of polynomial expressions in one variable, using a variety of strategies, including concrete objects and their related pictorial and symbolic models. b) Determine the product of polynomial expressions in one variable, using a variety of strategies, including concrete objects and their related pictorial and symbolic models, the application of the distributive property, and the use of area models. The factors should be limited to five or fewer terms. c) Factor completely first- and second-degree polynomials in one variable with integral coefficients. After factoring out the greatest common factor (GCF), leading coefficients should have no more than four factors. d) Determine the quotient of polynomials, using a monomial or binomial divisor, or a completely factored divisor. e) Represent and demonstrate equality of quadratic expressions in different forms.

By the end of this chapter you will be able to:

• Divide a polynomial by a monomial, term by term, using the quotient law from Chapter 10 (A.EO.2d)
• Divide a polynomial by a binomial with an area model and symbolically, then check by multiplying the quotient by the divisor (A.EO.2d)
• Divide by a completely factored divisor by cancellation, and state the domain restriction the cancellation forces (A.EO.2d)
• Rewrite $x^{2} + bx$ by completing the square, as an expression rewrite — not as a method for solving equations (A.EO.2e)
• Write one quadratic in standard form, factored form, and vertex form, and prove the three are the same expression by expanding (A.EO.2e)
• Say what each form reveals about the expression — the constant and leading coefficient, the inputs that make it zero, and the least or greatest value — without graphing a parabola (A.EO.2e)

Lessons: 14.1 Dividing by a Monomial · 14.2 Dividing by a Binomial · 14.3 Completely Factored Divisors and Domain Restrictions · 14.4 Equivalent Quadratic Forms

Why this chapter matters. Chapter 12 taught you to multiply polynomials, and Chapter 13 taught you to factor them. Division is the third operation, and it is the same product picture read as "one side and the area are known; find the other side." That skill is what lets you simplify a ratio of polynomials when a factor is shared. The second half of the chapter is different in look but the same in spirit: one quadratic expression can be written three ways, and A.EO.2e asks you to demonstrate that the three writings are equal. Expanding proves it. Completing the square produces the vertex form. Factored form comes from Chapter 13. None of that requires a parabola — the graph of a quadratic function is Chapter 16 — and none of it solves an equation. Solving by completing the square is Chapter 15.

Scope note. A.EO.2d limits divisors to a monomial, a binomial, or a completely factored divisor. No item in this chapter divides by an unfactored trinomial, and no item asks for polynomial long division beyond what an area model or cancellation can finish. A.EO.2e asks for equality of quadratic expressions in different forms. The three forms this chapter uses are standard $ax^{2} + bx + c$, factored $a(x - r)(x - s)$, and vertex $a(x - h)^{2} + k$. Completing the square appears here as an expression rewrite that produces vertex form; using it to solve $ax^{2} + bx + c = 0$ is A.EI.3 in Chapter 15. Nothing here graphs a parabola, names an axis of symmetry, or describes a transformation of $y = x^{2}$ — those are A.F.2 in Chapter 16. Factoring a polynomial from scratch is Chapter 13; this chapter uses factored form when it is already in hand or when equality needs proving by expanding.

Conventions this chapter fixes.

• A quotient is the result of a division. In $\dfrac{P(x)}{D(x)} = Q(x)$, the polynomial $P$ is the dividend, $D$ is the divisor, and $Q$ is the quotient.
• Every cancellation carries a domain restriction. Canceling a factor $(x - a)$ from numerator and denominator is valid only for $x \neq a$, because the original expression is undefined at $x = a$. The simplified expression and the original agree everywhere the original exists; they do not agree at the hole. Reporting the restriction is part of the answer.
• A monomial divisor $cx^{n}$ with $n \ge 1$ forces $x \neq 0$. The chapter states that restriction whenever the divisor contains a variable power.
• Vertex form $a(x - h)^{2} + k$ is an algebraic rewriting. The number $k$ is the least value of the expression when $a > 0$, and the greatest value when $a < 0$; it is reached at $x = h$. That claim is about the expression's values, not about a drawn curve.
• Proving two forms equal means expanding until both match, or rewriting one into the other by completing the square. Evaluating both at a few inputs is useful evidence and a fine use of technology, but it is not a proof — agreement at seven inputs does not rule out disagreement at an eighth.
• Item numbering runs straight through the chapter, from 1 in Lesson 14.1 to 110 at the end of the review. It does not restart at each lesson.

Calculator note. Algebra 1 has no no-calculator standards, and the Desmos Virginia calculator is available for the entire End-of-Course test. Use it here the way Chapter 12 does: pick an allowed input, evaluate the original expression and your quotient (or both quadratic forms), and confirm the numbers agree. For a division with a restriction, never choose the forbidden input — the original expression does not exist there. For equality of forms, a table of matching outputs is strong evidence; expanding is the proof.

Lesson 14.1 — Dividing by a Monomial

One dividend, one fraction per term

A.EO.2d begins with the simplest allowed divisor: a monomial. The procedure is the quotient law from Chapter 10, applied once per term.

$$\dfrac{12x^{4} + 18x^{3} - 6x^{2}}{3x^{2}} = \dfrac{12x^{4}}{3x^{2}} + \dfrac{18x^{3}}{3x^{2}} - \dfrac{6x^{2}}{3x^{2}}$$

Each separate division divides the coefficients and subtracts the exponents. The three quotients are $4x^{2}$, $6x$, and $-2$, so

$$\dfrac{12x^{4} + 18x^{3} - 6x^{2}}{3x^{2}} = 4x^{2} + 6x - 2$$

Read the figure one row at a time.

• Coefficients divide: $12 \div 3 = 4$, $18 \div 3 = 6$, $-6 \div 3 = -2$.
• Exponents subtract: $4 - 2 = 2$, $3 - 2 = 1$, $2 - 2 = 0$. An exponent of $0$ means the variable disappears and the term is a constant.
• The rewriting is valid for every $x \neq 0$, because the original expression has $3x^{2}$ in the denominator and does not exist at $x = 0$.

The procedure in full

1. Write the dividend as a sum of separate fractions, one for each term, all with the same monomial denominator.
2. Divide the coefficients in each fraction, signs included.
3. Subtract the exponents on the variable in each fraction, reading a bare $x$ as $x^{1}$.
4. State the restriction forced by the divisor: if the divisor contains $x^{n}$ with $n \ge 1$, then $x \neq 0$.

The same work in reverse is Chapter 12's monomial multiplication. Checking $\dfrac{12x^{4} + 18x^{3} - 6x^{2}}{3x^{2}} = 4x^{2} + 6x - 2$ means verifying

$$3x^{2}(4x^{2} + 6x - 2) = 12x^{4} + 18x^{3} - 6x^{2}$$

which it does, for every $x$ — and the original division is defined for every $x \neq 0$.

Signs and missing terms

Carry each sign with its term. In $\dfrac{8x^{3} - 12x^{2} + 4x}{4x}$ the middle term is $-12x^{2}$, so the second fraction is $\dfrac{-12x^{2}}{4x} = -3x$. Writing the middle term as if it were positive is the single most common error in this lesson.

A dividend may skip a power. $\dfrac{15x^{5} + 5x}{5x}$ has no $x^{3}$ or $x^{2}$ term, and that is fine — write two fractions, not five. The quotient is $3x^{4} + 1$, for $x \neq 0$.

Worked examples

Example 1 — The figure's division

Divide $12x^{4} + 18x^{3} - 6x^{2}$ by $3x^{2}$, and state the restriction.

Split into three fractions. Coefficients: $12 \div 3 = 4$, $18 \div 3 = 6$, $-6 \div 3 = -2$. Exponents: $4 - 2 = 2$, $3 - 2 = 1$, $2 - 2 = 0$.

Answer: $4x^{2} + 6x - 2$, for $x \neq 0$.

Example 2 — A linear monomial divisor

Divide $8x^{3} - 12x^{2} + 4x$ by $4x$.

$$\dfrac{8x^{3}}{4x} - \dfrac{12x^{2}}{4x} + \dfrac{4x}{4x} = 2x^{2} - 3x + 1$$

Answer: $2x^{2} - 3x + 1$, for $x \neq 0$.

Example 3 — A negative leading term

Divide $-20x^{6} + 10x^{4} - 30x^{2}$ by $5x^{2}$.

$$\dfrac{-20x^{6}}{5x^{2}} + \dfrac{10x^{4}}{5x^{2}} - \dfrac{30x^{2}}{5x^{2}} = -4x^{4} + 2x^{2} - 6$$

Answer: $-4x^{4} + 2x^{2} - 6$, for $x \neq 0$.

Example 4 — Checking by multiplying back

Check Example 2 by multiplying $4x$ by $2x^{2} - 3x + 1$.

$$4x(2x^{2} - 3x + 1) = 8x^{3} - 12x^{2} + 4x$$

Answer: The product recovers the dividend, so the quotient is correct wherever the original exists.

Example 5 — A constant monomial

Divide $6x^{2} - 9x + 3$ by $3$.

A constant divisor imposes no variable restriction. Each coefficient simply divides by $3$.

Answer: $2x^{2} - 3x + 1$. (No restriction from the divisor.)

Guided practice

1. Use the monomial-divisor figure. Write the three separate fractions it shows for $\dfrac{12x^{4} + 18x^{3} - 6x^{2}}{3x^{2}}$.
2. In that same figure, complete the first row of the table: coefficients divide, exponents subtract, quotient term.
3. In that same figure, why does the third row produce the constant $-2$ rather than a power of $x$?
4. In that same figure, state the restriction and explain why it is required.
5. Divide $15x^{5} - 10x^{3} + 5x$ by $5x$, showing one fraction per term, and state the restriction.
6. Check your answer to item 5 by multiplying the quotient by $5x$.

Independent practice

7. Divide. State the restriction when the divisor contains a variable. a) $\dfrac{18x^{4} + 27x^{3} - 9x^{2}}{9x^{2}}$ b) $\dfrac{10x^{3} - 15x^{2} + 5x}{5x}$ c) $\dfrac{24x^{5} - 16x^{3}}{8x}$ d) $\dfrac{14x^{4} + 21x^{2}}{7x^{2}}$
8. Divide. a) $\dfrac{12x^{3} - 18x^{2} + 6x}{6x}$ b) $\dfrac{-20x^{6} + 10x^{4} - 30x^{2}}{5x^{2}}$ c) $\dfrac{9x^{4} - 3x^{2} + 6}{3}$ d) $\dfrac{16x^{5} - 4x^{3}}{4x^{3}}$
9. Divide $\dfrac{x^{4} - 5x^{3} + 2x}{x}$ and state the restriction.
10. Divide $\dfrac{25x^{4} - 10x^{3} + 15x^{2}}{5x^{2}}$ and state the restriction.
11. Reasoning. Explain why $\dfrac{12x^{4} + 18x^{3} - 6x^{2}}{3x^{2}}$ and $4x^{2} + 6x - 2$ are not identical as expressions, even though they agree at every $x \neq 0$.
12. Error analysis. A student writes $\dfrac{8x^{3} - 12x^{2} + 4x}{4x} = 2x^{2} - 3x + 4x$. Identify the error and give the correct quotient.
13. Error analysis. A student writes $\dfrac{15x^{5} - 10x^{3}}{5x} = 3x^{4} - 2x^{3}$. Identify the error and give the correct quotient with its restriction.
14. Check $\dfrac{18x^{4} + 27x^{3} - 9x^{2}}{9x^{2}} = 2x^{2} + 3x - 1$ by multiplying back. Show the product.
15. Application. The volume of a rectangular box is $24x^{3} + 36x^{2}$ cubic centimeters, and its height is $12x$ centimeters. Write an expression for the area of the base, and state any restriction $x$ must obey for the model to make sense.
16. Evaluate both $\dfrac{12x^{4} + 18x^{3} - 6x^{2}}{3x^{2}}$ and $4x^{2} + 6x - 2$ at $x = 2$. Confirm they agree, and explain why you may not check at $x = 0$.
17. Divide $\dfrac{-8x^{4} + 12x^{3} - 4x^{2}}{-4x^{2}}$ and simplify the signs carefully.
18. Write a monomial dividend, a monomial divisor, and their quotient so that the quotient is $5x^{2} - 3$, with restriction $x \neq 0$.
19. Divide $\dfrac{27x^{6} - 18x^{4} + 9x^{2}}{9x^{2}}$.
20. Reasoning. Why does dividing a polynomial by a constant monomial require no restriction of the form $x \neq 0$, while dividing by $3x^{2}$ does?

Exit ticket 14.1

21. Divide $\dfrac{12x^{4} + 18x^{3} - 6x^{2}}{3x^{2}}$ and state the restriction.
22. Divide $\dfrac{20x^{5} - 15x^{3} + 10x}{5x}$.
23. Check your answer to item 22 by multiplying the quotient by $5x$.
24. Explain, in one sentence, why the quotient law subtracts exponents rather than dividing them.

Lesson 14.2 — Dividing by a Binomial

Division as a missing side

Chapter 12 filled an area model forwards: two sides given, the interior cells the product. Division runs the same grid backwards. The total area is the dividend, one side is the divisor, and the missing side is the quotient.

The figure sets up $\dfrac{2x^{2} + 7x + 3}{x + 3}$. The top edge is already labeled $x$ and $+3$. The four cells add to the dividend: $2x^{2} + 6x + x + 3 = 2x^{2} + 7x + 3$. Each left label is forced by dividing a cell by the top label above it:

• Top row: $2x^{2} \div x = 2x$
• Bottom row: $3 \div 3 = 1$

So the quotient is $2x + 1$. The check is Chapter 12 run forwards:

$$(x + 3)(2x + 1) = 2x^{2} + 6x + x + 3 = 2x^{2} + 7x + 3$$

The multiply-back check

Every binomial division in this chapter gets the same self-check. For $\dfrac{6x^{2} + 7x - 3}{3x - 1} = 2x + 3$:

$$(3x - 1)(2x + 3) = 6x^{2} + 9x - 2x - 3 = 6x^{2} + 7x - 3$$

If the product recovers the dividend, the quotient is correct wherever the divisor is not zero. The restriction here is $x \neq \tfrac13$, because the original expression is undefined when $3x - 1 = 0$.

Finding the quotient without a picture

When the dividend is a quadratic and the divisor is linear, the quotient is linear. You can find it by asking what binomial times the divisor recovers the dividend — which is Chapter 13's factoring, read as division — or by filling the area model. Both routes are allowed; both end at the multiply-back check.

For $\dfrac{x^{2} + 9x + 20}{x + 4}$:

• The factors of $x^{2} + 9x + 20$ are $(x + 4)(x + 5)$, so the quotient is $x + 5$, for $x \neq -4$.
• Or fill the area model: top edge $x$ and $+4$; left edge forced as $x$ and $+5$; cells $x^{2}$, $4x$, $5x$, $20$.

Worked examples

Example 1 — The area-model figure

Use the area-model figure to divide $2x^{2} + 7x + 3$ by $x + 3$.

Left labels: $2x^{2} \div x = 2x$ and $3 \div 3 = 1$.

Answer: $2x + 1$, for $x \neq -3$.

Example 2 — Multiply back

Check Example 1 by multiplying $(x + 3)(2x + 1)$.

Answer: $2x^{2} + 7x + 3$, matching the dividend.

Example 3 — A leading coefficient other than 1

Divide $6x^{2} + 7x - 3$ by $3x - 1$, and state the restriction.

The quotient is $2x + 3$, because $(3x - 1)(2x + 3) = 6x^{2} + 7x - 3$.

Answer: $2x + 3$, for $x \neq \tfrac13$.

Example 4 — Difference of squares

Divide $x^{2} - 16$ by $x - 4$.

$x^{2} - 16 = (x - 4)(x + 4)$, so the quotient is $x + 4$.

Answer: $x + 4$, for $x \neq 4$.

Example 5 — Blank area models

The blank area-model figure poses three divisions. Find each quotient.

• $\dfrac{x^{2} + 9x + 20}{x + 4} = x + 5$, for $x \neq -4$
• $\dfrac{2x^{2} + 11x + 12}{x + 4} = 2x + 3$, for $x \neq -4$
• $\dfrac{3x^{2} - 5x - 2}{3x + 1} = x - 2$, for $x \neq -\tfrac13$

Answer: $x + 5$; $2x + 3$; $x - 2$, each with its restriction.

Guided practice

25. Use the area-model figure. Name the divisor on the top edge and the four cells that add to the dividend $2x^{2} + 7x + 3$.
26. In that same figure, explain how the left-edge labels $2x$ and $+1$ are forced.
27. Use the multiply-back figure. Write the product $(3x - 1)(2x + 3)$ and confirm it recovers $6x^{2} + 7x - 3$.
28. Divide $x^{2} + 5x + 6$ by $x + 2$ using an area model or by factoring, and state the restriction.
29. Divide $x^{2} - 9$ by $x + 3$, and state the restriction.
30. Why must the restriction for $\dfrac{6x^{2} + 7x - 3}{3x - 1}$ be $x \neq \tfrac13$ rather than $x \neq 0$?

Independent practice

31. Divide. State the restriction. a) $\dfrac{x^{2} + 9x + 20}{x + 4}$ b) $\dfrac{x^{2} - 5x + 6}{x - 2}$ c) $\dfrac{x^{2} + 5x + 6}{x + 2}$ d) $\dfrac{x^{2} - 16}{x - 4}$
32. Divide. a) $\dfrac{2x^{2} + 7x + 3}{x + 3}$ b) $\dfrac{2x^{2} + 11x + 12}{x + 4}$ c) $\dfrac{6x^{2} + 7x - 3}{3x - 1}$ d) $\dfrac{3x^{2} - 5x - 2}{3x + 1}$
33. Divide $\dfrac{2x^{2} - x - 3}{x + 1}$ and check by multiplying back.
34. Divide $\dfrac{4x^{2} - 4x - 3}{2x + 1}$ and state the restriction.
35. Use the blank area-model figure. Fill all three models: write the top-edge labels, the left-edge quotient, and the four cells for each.
36. In the blank area-model figure, check the middle model by adding the four cells and confirming they total $2x^{2} + 11x + 12$.
37. Application. A rectangular garden has area $x^{2} + 9x + 20$ square meters and width $x + 4$ meters. Write an expression for its length, and state any restriction.
38. Application. A store's weekly revenue is $(3x - 1)(2x + 3)$ dollars when it sells $x$ cases of an item. Expand the revenue into a single polynomial. Then divide that polynomial by $3x - 1$ and state the restriction.
39. Error analysis. A student claims $\dfrac{x^{2} + 9x + 20}{x + 4} = x + 4$. Identify the error, give the correct quotient, and show the multiply-back check.
40. Error analysis. A student divides $2x^{2} + 7x + 3$ by $x + 3$ and gets $2x + 3$. Multiply back and show why that quotient is wrong; then give the correct one.
41. Divide $\dfrac{x^{2} - 8x + 15}{x - 3}$.
42. Divide $\dfrac{5x^{2} + 14x - 3}{x + 3}$.
43. Check $\dfrac{3x^{2} - 5x - 2}{3x + 1} = x - 2$ by multiplying $(3x + 1)(x - 2)$.
44. Reasoning. Explain why an area model for division needs the cells to add to the dividend, not multiply to it.
45. Evaluate both $\dfrac{2x^{2} + 7x + 3}{x + 3}$ and $2x + 1$ at $x = 1$. Confirm they agree.
46. Divide $\dfrac{x^{2} + 7x + 12}{x + 3}$ and state the restriction.

Exit ticket 14.2

47. Divide $\dfrac{2x^{2} + 7x + 3}{x + 3}$ and state the restriction.
48. Divide $\dfrac{6x^{2} + 7x - 3}{3x - 1}$ and check by multiplying back.
49. Using an area model or factoring, divide $\dfrac{x^{2} + 9x + 20}{x + 4}$.
50. Why does multiplying the quotient by the divisor check a division?

Lesson 14.3 — Completely Factored Divisors and Domain Restrictions

Cancellation is not optional about its restriction

A.EO.2d's third allowed divisor is a completely factored one. When a factor appears in both the numerator and the denominator, it cancels — and the cancellation is valid only where that factor is not zero.

The figure walks through

$$\dfrac{(x - 3)(x + 1)}{x - 3} = \dfrac{x - 3}{x - 3} \cdot (x + 1) = 1 \cdot (x + 1) = x + 1 \quad \text{provided } x \neq 3$$

At $x = 3$ the left side is $\dfrac{0 \cdot 4}{0}$, which is undefined, while the right side is $4$. The two expressions agree at every input except $3$, and at $3$ only one of them exists. So the quotient is $x + 1$ together with its restriction, not $x + 1$ alone.

What "completely factored" allows

The divisor may be a single binomial factor, or a product of factors already written out. You may cancel any factor that appears in both, and you must record a restriction for each canceled factor.

$$\dfrac{(x + 3)(x - 2)(x + 1)}{(x + 3)(x - 2)} = x + 1 \quad \text{for } x \neq -3 \text{ and } x \neq 2$$

You do not expand the numerator or the denominator first. Expanding would hide the common factors and turn a one-step cancellation into a harder division.

Connecting back to Lesson 14.2

Every binomial quotient from Lesson 14.2 can be rewritten this way once the dividend is factored. $\dfrac{x^{2} + 9x + 20}{x + 4} = \dfrac{(x + 4)(x + 5)}{x + 4} = x + 5$ for $x \neq -4$. Cancellation and the area model are two readings of the same fact.

Worked examples

Example 1 — The cancellation figure

Simplify $\dfrac{(x - 3)(x + 1)}{x - 3}$ and state the restriction.

Cancel $x - 3$, which requires $x \neq 3$.

Answer: $x + 1$, for $x \neq 3$.

Example 2 — A shared binomial with a coefficient

Simplify $\dfrac{(2x - 1)(x + 4)}{2x - 1}$.

Answer: $x + 4$, for $x \neq \tfrac12$.

Example 3 — Two factors canceled

Simplify $\dfrac{(x + 3)(x - 2)(x + 1)}{(x + 3)(x - 2)}$.

Answer: $x + 1$, for $x \neq -3$ and $x \neq 2$.

Example 4 — A squared factor

Simplify $\dfrac{(x - 1)(x - 1)}{x - 1}$.

One factor of $x - 1$ cancels, leaving one factor of $x - 1$.

Answer: $x - 1$, for $x \neq 1$.

Example 5 — Why the restriction matters numerically

Show that $\dfrac{(x - 3)(x + 1)}{x - 3}$ and $x + 1$ disagree in status at $x = 3$, even though they agree at $x = 4$.

At $x = 4$: left side $\dfrac{1 \cdot 5}{1} = 5$, right side $5$. At $x = 3$: left side undefined, right side $4$.

Answer: They agree at $4$; at $3$ only the simplified expression exists.

Guided practice

51. Use the cancellation figure. Write the four steps it shows from $\dfrac{(x - 3)(x + 1)}{x - 3}$ to $x + 1$.
52. In that same figure, why is $\dfrac{x - 3}{x - 3}$ equal to $1$ only when $x \neq 3$?
53. In that same figure, evaluate both sides of the claimed equality at $x = 3$ and at $x = 4$. What happens at each?
54. Simplify $\dfrac{(x + 2)(x - 5)}{x + 2}$ and state the restriction.
55. Simplify $\dfrac{(x + 3)(x - 2)(x + 1)}{(x + 3)(x - 2)}$ and state both restrictions.
56. Rewrite $\dfrac{x^{2} + 9x + 20}{x + 4}$ by factoring the numerator first, then canceling. State the restriction.

Independent practice

57. Simplify. State every restriction. a) $\dfrac{(x - 3)(x + 1)}{x - 3}$ b) $\dfrac{(x + 2)(x - 5)}{x + 2}$ c) $\dfrac{(2x - 1)(x + 4)}{2x - 1}$ d) $\dfrac{(x - 4)(x + 4)}{x - 4}$
58. Simplify. a) $\dfrac{(x + 1)(x + 1)}{x + 1}$ b) $\dfrac{(x - 2)(x + 3)(x - 5)}{(x - 2)(x + 3)}$ c) $\dfrac{(3x + 1)(x - 2)}{3x + 1}$ d) $\dfrac{(x + 6)(2x - 3)}{x + 6}$
59. Factor the numerator, then simplify $\dfrac{x^{2} - 5x + 6}{x - 2}$.
60. Factor the numerator, then simplify $\dfrac{x^{2} - 16}{x + 4}$.
61. Reasoning. A student simplifies $\dfrac{(x - 3)(x + 1)}{x - 3}$ to $x + 1$ and stops. Explain what is missing from the answer and why it matters.
62. Error analysis. A student cancels the $x$ in $\dfrac{x + 5}{x}$ and writes $5$. Identify the error. (You may not cancel a term across a sum.)
63. Error analysis. A student writes $\dfrac{(x - 1)(x - 1)}{x - 1} = 1$. Identify the error and give the correct simplified form with its restriction.
64. Evaluate $\dfrac{(x - 3)(x + 1)}{x - 3}$ and $x + 1$ at $x = 0$ and at $x = 5$. Confirm agreement, and explain why $x = 3$ is not an allowed check.
65. Application. A company's profit density is modeled by $\dfrac{(x - 2)(x + 5)}{x - 2}$ dollars per unit when it produces $x$ hundred units. Simplify the expression and state the production level at which the model is undefined.
66. Simplify $\dfrac{(x + 4)(x - 1)(x + 2)}{(x + 4)(x + 2)}$.
67. Connect to Lesson 14.2: write $\dfrac{6x^{2} + 7x - 3}{3x - 1}$ by factoring the numerator as $(3x - 1)(2x + 3)$, then cancel.
68. Reasoning. Why does A.EO.2d insist the divisor be completely factored before you cancel, rather than allowing you to cancel pieces of an unfactored trinomial?
69. Simplify $\dfrac{(5x - 2)(x + 7)}{5x - 2}$.
70. Write an original rational expression whose simplified form is $x - 5$ with restriction $x \neq 2$.

Exit ticket 14.3

71. Simplify $\dfrac{(x - 3)(x + 1)}{x - 3}$ and state the restriction.
72. Simplify $\dfrac{(x + 3)(x - 2)(x + 1)}{(x + 3)(x - 2)}$.
73. Factor, then simplify $\dfrac{x^{2} + 5x + 6}{x + 2}$.
74. Explain why the simplified form and the original expression are not the same at the restricted input.

Lesson 14.4 — Equivalent Quadratic Forms

One expression, three writings

A.EO.2e asks you to represent and demonstrate equality of quadratic expressions in different forms. This chapter uses three:

The figure puts $x^{2} - 6x + 5$, $(x - 1)(x - 5)$, and $(x - 3)^{2} - 4$ side by side. Expanding proves the equality:

$$(x - 1)(x - 5) = x^{2} - 5x - x + 5 = x^{2} - 6x + 5$$

$$(x - 3)^{2} - 4 = x^{2} - 6x + 9 - 4 = x^{2} - 6x + 5$$

No parabola is required. The three writings are the same expression because algebra says so.

Completing the square produces vertex form

To rewrite $x^{2} + bx$ in vertex form:

1. Split the middle term into two equal strips of $\tfrac{b}{2}\,x$.
2. Add the missing corner $\left(\dfrac{b}{2}\right)^{2}$ so the figure becomes a square, $\left(x + \dfrac{b}{2}\right)^{2}$.
3. Subtract the same corner immediately, so the value of the expression does not change.

For $x^{2} + 8x$:

$$x^{2} + 8x = x^{2} + 8x + 16 - 16 = (x + 4)^{2} - 16$$

The area picture supplies the $16$; the $-16$ is what keeps the new expression equal to the old one. Check at $x = 1$: $1 + 8 = 9$, and $(1 + 4)^{2} - 16 = 25 - 16 = 9$.

For a full quadratic $x^{2} + bx + c$, complete the square on $x^{2} + bx$ and carry $c$ along:

$$x^{2} - 6x + 5 = (x^{2} - 6x + 9) - 9 + 5 = (x - 3)^{2} - 4$$

This chapter never sets the expression equal to zero and solves. Completing the square as a solution method is Chapter 15.

Tables give evidence; expanding gives proof

The table evaluates all three forms of $x^{2} - 6x + 5$ at $x = -1, 0, 1, 2, 3, 4, 5$. Every column agrees. That is strong evidence and a fine use of technology — but it is not proof. Seven agreements do not rule out an eighth disagreement. Expanding does: it shows the forms are the same expression, so they agree at every input.

Worked examples

Example 1 — Completing the square on $x^{2} + bx$

Rewrite $x^{2} + 8x$ in vertex form.

Half of $8$ is $4$; $4^{2} = 16$.

Answer: $(x + 4)^{2} - 16$

Example 2 — A negative middle term

Rewrite $x^{2} - 4x$ in vertex form.

Half of $-4$ is $-2$; $(-2)^{2} = 4$.

Answer: $(x - 2)^{2} - 4$

Example 3 — Full quadratic to vertex form

Rewrite $x^{2} - 6x + 5$ in vertex form.

$$x^{2} - 6x + 5 = (x^{2} - 6x + 9) - 9 + 5 = (x - 3)^{2} - 4$$

Answer: $(x - 3)^{2} - 4$

Example 4 — Proving equality by expanding

Show that $(x - 1)(x - 5)$ and $(x - 3)^{2} - 4$ both equal $x^{2} - 6x + 5$.

Answer: $(x - 1)(x - 5) = x^{2} - 6x + 5$ and $(x - 3)^{2} - 4 = x^{2} - 6x + 5$, as in the three-forms figure.

Example 5 — What each form reveals

For $x^{2} - 6x + 5 = (x - 1)(x - 5) = (x - 3)^{2} - 4$, state what each form reveals.

Answer: Standard reveals leading coefficient $1$ and constant $5$. Factored reveals the expression is $0$ at $x = 1$ and at $x = 5$. Vertex reveals the least value is $-4$, reached at $x = 3$.

Guided practice

75. Use the completing-the-square figure. How is the missing corner's area computed for $x^{2} + 8x$, and why is $-16$ written after $(x + 4)^{2}$?
76. Use the three-forms figure. Write the three expressions it shows, and expand the factored form to standard form.
77. In that same figure, expand the vertex form to standard form.
78. Use the table figure. At $x = 3$, what output do all three forms give, and what does that output tell you about the vertex form?
79. Rewrite $x^{2} + 6x$ by completing the square.
80. Rewrite $x^{2} + 10x$ by completing the square.
81. Rewrite $x^{2} - 4x$ by completing the square.
82. Show by expanding that $(x - 2)^{2} - 1 = x^{2} - 4x + 3$.

Independent practice

83. Rewrite each by completing the square. a) $x^{2} + 6x$ b) $x^{2} + 10x$ c) $x^{2} - 4x$ d) $x^{2} - 6x$
84. Use the blank completing-the-square frames. Fill all three: the corner added and the vertex form.
85. Rewrite each quadratic in vertex form. a) $x^{2} - 6x + 5$ b) $x^{2} - 4x + 3$ c) $x^{2} + 2x - 8$ d) $x^{2} - 8x + 12$
86. Expand each to standard form, proving equality. a) $(x - 1)(x - 5)$ b) $(x - 3)^{2} - 4$ c) $(x + 4)(x - 2)$ d) $(x + 1)^{2} - 9$
87. For $x^{2} - 4x + 3$, write the factored form and the vertex form, then expand both to prove they match the standard form.
88. Use the blank three-form panels. For $x^{2} - 6x + 5$, fill standard, factored, and vertex forms, note what each reveals, and write the expansion that proves equality.
89. Application. A ball's height, in feet, $t$ seconds after being thrown is modeled by the expression $-16t^{2} + 32t + 48$. Rewrite $-16t^{2} + 32t$ by factoring out $-16$ and completing the square on $t^{2} - 2t$, then combine the constant. What is the greatest value of the height expression, and at what $t$ is it reached? (Do not solve an equation; read $h$ and $k$ from vertex form.)
90. Build a table for $x^{2} + 2x - 8$, $(x + 4)(x - 2)$, and $(x + 1)^{2} - 9$ at $x = -4, -1, 0, 2$. Confirm the three rows agree, then expand to prove equality.
91. Reasoning. Why is a table of matching outputs evidence but not proof that two forms are equal?
92. Error analysis. A student rewrites $x^{2} + 8x$ as $(x + 4)^{2}$ and stops. Identify the error and give the correct vertex form.
93. Error analysis. A student claims $x^{2} - 6x + 5$ and $(x - 3)^{2} + 4$ are equal because both involve $3$ and $4$. Expand the second expression and show they are not equal; then give the correct vertex form.
94. For $2(x - 2)^{2} - 2$, expand to standard form. Then state the least value of the expression and the input at which it occurs.
95. Write $x^{2} - 8x + 12$ in all three forms and say what each reveals.
96. Reasoning. Vertex form reveals a least or greatest value of an expression. Explain how you know, from $a(x - h)^{2} + k$ alone with $a > 0$, that the least value is $k$ at $x = h$ — without drawing a parabola.

Exit ticket 14.4

97. Rewrite $x^{2} + 8x$ in vertex form.
98. Rewrite $x^{2} - 6x + 5$ in all three forms.
99. Expand $(x - 3)^{2} - 4$ and $(x - 1)(x - 5)$ to prove both equal $x^{2} - 6x + 5$.
100. What does the vertex form $(x - 3)^{2} - 4$ reveal that the standard form does not?

Chapter 14 Review

Vocabulary. dividend · divisor · quotient · monomial divisor · binomial divisor · completely factored divisor · domain restriction · cancellation · area model · multiply-back check · standard form · factored form · vertex form · completing the square · equivalent forms

A.EO.2 d and e ask for two different skills, so this review is organized by both. Part A is monomial division, Part B is binomial division and the multiply-back check, Part C is cancellation with restrictions, Part D is completing the square and the three forms, and Part E mixes them.

Part A — Monomial divisors

101. Divide $\dfrac{12x^{4} + 18x^{3} - 6x^{2}}{3x^{2}}$ and state the restriction.
102. Divide $\dfrac{-20x^{6} + 10x^{4} - 30x^{2}}{5x^{2}}$.
103. Check your answer to item 101 by multiplying the quotient by $3x^{2}$.

Part B — Binomial divisors

104. Divide $\dfrac{2x^{2} + 7x + 3}{x + 3}$ and state the restriction.
105. Divide $\dfrac{6x^{2} + 7x - 3}{3x - 1}$ and check by multiplying back.
106. Divide $\dfrac{x^{2} + 9x + 20}{x + 4}$ using an area model or by factoring.

Part C — Completely factored divisors

107. Simplify $\dfrac{(x - 3)(x + 1)}{x - 3}$ and state the restriction.
108. Simplify $\dfrac{(x + 3)(x - 2)(x + 1)}{(x + 3)(x - 2)}$ and state both restrictions.

Part D — Equivalent quadratic forms

109. Rewrite $x^{2} - 6x + 5$ in factored form and in vertex form, then expand both to prove equality with the standard form.
110. Rewrite $x^{2} + 10x$ by completing the square. State the least value of the expression and the input at which it occurs.

Standards coverage check — Chapter 14

Supporting items: 11, 20, 30, 44, 61, 68, 74, 91, and 96 are reasoning prompts aimed at the claims most often taken on faith — why a restriction is part of the answer, why expanding proves equality, and why vertex form reveals a least value algebraically. Items 12, 13, 39, 40, 62, 63, 92, and 93 are error analyses aimed at the most common defects: dropping a sign, dividing exponents, canceling terms across a sum, and forgetting to subtract the corner when completing the square.

Boundaries respected. Every divisor in this chapter is a monomial, a binomial, or a completely factored expression — never an unfactored trinomial. Completing the square is used only to rewrite expressions into vertex form; no item solves $ax^{2} + bx + c = 0$ by completing the square, by the quadratic formula, or by the zero product property — that is A.EI.3 in Chapter 15. No item graphs a parabola, names an axis of symmetry, or describes a transformation of $y = x^{2}$ — that is A.F.2 in Chapter 16. Factoring a polynomial from scratch is practiced only when a numerator must be factored in order to cancel; the full factoring skill is A.EO.2c in Chapter 13. Every domain restriction created by a cancellation or a variable divisor is stated with the answer.

Answer keys for every item in this chapter are in Appendix A.