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9.5: Factoring Quadratic Expressions

Created by: CK-12

Learning Objectives

  • Write quadratic equations in standard form.
  • Factor quadratic expressions for different coefficient values.
  • Factor when a = -1.

Write Quadratic Expressions in Standard Form

Quadratic polynomials are polynomials of 2^{nd} degree. The standard form of a quadratic polynomial is written as

ax^2 + bx + c

Here a, b, and c stand for constant numbers. Factoring these polynomials depends on the values of these constants. In this section, we will learn how to factor quadratic polynomials for different values of a, b, and c. In the last section, we factored common monomials, so you already know how to factor quadratic polynomials where c = 0.

For example for the quadratic ax^2 + bx, the common factor is x and this expression is factored as x(ax + b). When all the coefficients are not zero these expressions are also called Quadratic Trinomials, since they are polynomials with three terms.

Factor when a = 1, b is Positive, and c is Positive

Let’s first consider the case where a = 1, b is positive and c is positive. The quadratic trinomials will take the following form.

x^2 + bx + c

You know from multiplying binomials that when you multiply two factors (x + m) (x + n) you obtain a quadratic polynomial. Let’s multiply this and see what happens. We use The Distributive Property.

(x + m) (x + n) = x^2 + nx + mx + mn

To simplify this polynomial we would combine the like terms in the middle by adding them.

(x + m) (x + n) = x^2 + (n + m)x + mn

To factor we need to do this process in reverse.

& \text{We see that} && x^2 + (n + m) x + mn \\& \text{Is the same form as} && x^2 + bx + c

This means that we need to find two numbers m and n where

n + m = b \quad \quad \text{and} \quad \quad mn = c

To factor x^2 + bx + c, the answer is the product of two parentheses.

(x + m) (x + n)

so that n + m = b and mn = c

Let’s try some specific examples.

Example 1

Factor x^2 + 5x +6

Solution We are looking for an answer that is a product of two binomials in parentheses.

(x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;})

To fill in the blanks, we want two numbers m and n that multiply to 6 and add to 5. A good strategy is to list the possible ways we can multiply two numbers to give us 6 and then see which of these pairs of numbers add to 5. The number six can be written as the product of.

&6 = 1 \cdot 6 && \text{and} && 1 + 6 = 7 \\&6 = 2 \cdot 3 && \text{and} && 2 + 3 = 5 \qquad \leftarrow \qquad \text{This is the correct choice.}

So the answer is (x + 2) (x + 3).

We can check to see if this is correct by multiplying (x + 2) (x + 3).

& \quad \quad \quad x \ \ + \  2\\& \underline{\;\;\;\;\;\;\;\;\;\;\;x \ \ + \  3}\\& \quad \quad \quad 3x \ + \ 6\\& \underline{x^2 \ + \ 2x\;\;\;\;\;\;\;\;\;}\\& x^2 \ + \  5x \ + \ 9

The answer checks out.

Example 2

Factor x ^2 + 7x + 12

Solution

We are looking for an answer that is a product of two parentheses (x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;}).

The number 12 can be written as the product of the following numbers.

& 12 = 1 \cdot 12  && \text{and} && 1 + 12 = 13 \\& 12 = 2 \cdot 6 && \text{and} &&  2 + 6 = 8 \\& 12 = 3 \cdot 4 && \text{and} &&  3 + 4 = 7 \qquad \leftarrow \qquad \text{This is the correct choice.}

The answer is (x + 3) (x + 4).

Example 3

Factor x^2 + 8x + 12.

Solution

We are looking for an answer that is a product of the two parentheses (x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;}).

The number 12 can be written as the product of the following numbers.

& 12  = 1 \cdot 12 && \text{and} && 1 + 12 = 13 \\& 12  = 2 \cdot 6 && \text{and} && 2 + 6 = 8 \qquad \leftarrow \qquad \text{This is the correct choice}.\\& 12  = 3 \cdot 4 && \text{and} && 3 + 4 = 7

The answer is (x + 2) (x + 6).

Example 4

Factor x^2 + 12x + 36.

Solution

We are looking for an answer that is a product of the two parentheses (x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;}).

The number 36 can be written as the product of the following numbers.

& 36 = 1 \cdot 36 && \text{and} && 1 + 36 = 37 \\& 36 = 2 \cdot 18 && \text{and} && 2 + 18 = 20 \\& 36 = 3 \cdot 12 && \text{and} && 3 + 12 = 15 \\& 36 = 4 \cdot 9 && \text{and} && 4 + 9 = 13 \\& 36 = 6 \cdot 6 && \text{and} && 6 + 6 = 12 \qquad \leftarrow \qquad \text{This is the correct choice}

The answer is (x + 6) (x + 6).

Factor when a = 1, b is Negative and c is Positive

Now let’s see how this method works if the middle coefficient (b) is negative.

Example 5

Factor x^2 - 6x + 8.

Solution

We are looking for an answer that is a product of the two parentheses (x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;}).

The number 8 can be written as the product of the following numbers.

8 = 1 \cdot 8 and 1 + 8 = 9 Notice that these are two different choices.

But also,

& 8 = (-1) \cdot (-8) && \text{and} && -1 + (-8) = -9 && \text{Notice that these are two different choices.}\\& 8 = 2 \cdot 4 && \text{and} && 2 + 4 = 6

But also,

8 = (-2) \cdot (-4) && \text{and} && -2 + (-4) = -6 \qquad \leftarrow \qquad \text{This is the correct choice.}

The answer is (x - 2) (x - 4)

We can check to see if this is correct by multiplying (x - 2) (x - 4).

& \quad \quad \quad x \ \ - \  2\\& \underline{\;\;\;\;\;\;\;\;\;\;\;x \ \ - \  4}\\& \quad \  - \ 4x \ + \ 8 \\& \underline{x^2 \ - \ 2x\;\;\;\;\;\;\;\;\;}\\& x^2 \ - \  6x \ + \ 8

The answer checks out.

Example 6

Factor x - 17x + 16

Solution

We are looking for an answer that is a product of two parentheses: (x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;}).

The number 16 can be written as the product of the following numbers:

16 & = 1 \cdot 16 && \text{and} && 1 + 16 = 17 \\16 & = (-1) \cdot (-16) && \text{and} && -1 + (-16) = -17  \qquad \leftarrow \qquad  \text{This is the correct choice.}\\16 & = 2 \cdot 8 && \text{and} && 2 + 8 = 10 \\16 & = (-2) \cdot (-8) && \text{and} && -2 + (-8) = -10\\16 & = 4 \cdot 4 && \text{and} && 4 + 4 = 8 \\16 & = (-4) \cdot (-4) && \text{and} && -4 + (-4) = -8

The answer is (x - 1) (x - 16).

Factor when a = 1 and c is Negative

Now let’s see how this method works if the constant term is negative.

Example 7

Factor x^2 + 2x -15

Solution

We are looking for an answer that is a product of two parentheses (x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;}).

In this case, we must take the negative sign into account. The number -15 can be written as the product of the following numbers.

-15 = -1 \cdot 15 && \text{and} && -1 + 15 = 14 && \ \text{Notice that these are two different choices.}

And also,

-15 = 1 \cdot (-15) && \text{and} && 1 + (-15) = -14 && \text{Notice that these are two different choices.}

& -15 = -3 \cdot 5  && \text{and} && -3 + 5 = 2  \qquad \leftarrow \qquad \text{This is the correct choice.} \\& -15  = 3 \cdot (-5) && \text{and} && 3 + (-5) = -2

The answer is (x - 3) (x + 5).

We can check to see if this is correct by multiplying (x - 3) (x + 5).

& \quad \quad \quad \ x \ \ - \  3\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;x \ \ + \ 5}\\& \quad \quad \quad 5x \ - \ 15\\& \underline{x^2 \ -\ 3x\;\;\;\;\;\;\;\;\;\;}\\& x^2 \ + \  2x \ - \ 15

The answer checks out.

Example 8

Factor x^2 - 10x -24

Solution

We are looking for an answer that is a product of two parentheses (x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;}).

The number -24 can be written as the product of the following numbers.

& -24 = -1 \cdot 24 && \text{and} && -1 + 24 = 23 \\& -24 = 1 \cdot (-24) && \text{and} && 1 + (-24) = -23 \\& -24 = -2 \cdot 12 && \text{and} && -2 + 12 = 10 \\& -24 = 2 \cdot (-12) && \text{and} && 2 + (-12) = -10  \qquad \leftarrow \qquad \text{This is the correct choice.}\\& -24 = -3 \cdot 8 && \text{and} && -3 + 8 = 5 \\& -24 = 3 \cdot (-8) && \text{and} && 3 + (-8) = -5 \\& -24 = -4 \cdot 6 && \text{and} && -4 + 6 = 2 \\& -24 = 4 \cdot (-6) && \text{and} && 4 + (-6) = -2

The answer is (x - 12)(x + 2).

Example 9

Factor x^2 + 34x - 35

Solution

We are looking for an answer that is a product of two parentheses (x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})

The number -35 can be written as the product of the following numbers:

& -35 = -1 \cdot 35 && \text{and} && -1 + 35 = 34  \qquad \leftarrow \qquad \text{This is the correct choice.}\\& -35 = 1 \cdot (-35) && \text{and} && 1 + (-35) = -34 \\& -35 = -5 \cdot 7 && \text{and} &&  -5 + 7 = 2 \\& -35 = 5 \cdot (-7) && \text{and} && 5 + (-7) = -2

The answer is (x - 1) (x + 35).

Factor when a = - 1

When a = -1, the best strategy is to factor the common factor of -1 from all the terms in the quadratic polynomial. Then, you can apply the methods you have learned so far in this section to find the missing factors.

Example 10

Factor x^2 + x + 6.

Solution

First factor the common factor of -1 from each term in the trinomial. Factoring -1 changes the signs of each term in the expression.

- x^2 + x + 6 = - (x^2 - x - 6)

We are looking for an answer that is a product of two parentheses (x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})

Now our job is to factor x^2 - x - 6.

The number -6 can be written as the product of the following numbers.

& -6 = -1 \cdot 6 && \text{and}&&  -1 + 6 = 5 \\& -6 = 1 \cdot (-6) && \text{and}&& 1 + (-6) = -5 \\& -6 = -2 \cdot 3 && \text{and}&& -2 + 3 = 1 \\& -6 = 2 \cdot (-3) && \text{and}&& 2 + (-3) = -1  \qquad \leftarrow \qquad \text{This is the correct choice.}

The answer is - (x - 3) (x + 2).

To Summarize,

A quadratic of the form x + bx + c factors as a product of two parenthesis (x + m) (x + n).

  • If b and c are positive then both m and n are positive
    • Example x^2 + 8x + 12 factors as (x + 6) (x + 2).
  • If b is negative and c is positive then both m and n are negative.
    • Example x^2 - 6x + 8 factors as (x - 2) (x - 4).
  • If c is negative then either m is positive and n is negative or vice-versa
    • Example x^2 + 2x -15 factors as (x + 5) (x - 3).
    • Example x^2 + 34x -35 factors as (x + 35) (x - 1).
  • If a = -1, factor a common factor of -1 from each term in the trinomial and then factor as usual. The answer will have the form - (x + m) (x + n).
    • Example - x^2 + x + 6 factors as -(x - 3) (x + 2).

Review Questions

Factor the following quadratic polynomials.

  1. x^2 + 10x + 9
  2. x^2 + 15x + 50
  3. x^2 + 10x + 21
  4. x^2 + 16x + 48
  5. x^2 - 11x + 24
  6. x^2 - 13x + 42
  7. x^2 - 14x +33
  8. x^2 - 9x + 20
  9. x^2 + 5x - 14
  10. x^2 + 6x - 27
  11. x^2 + 7x - 78
  12. x^2 + 4x - 32
  13. x^2 - 12x - 45
  14. x^2 - 5x - 50
  15. x^2 - 3x - 40
  16. x^2 - x - 56
  17. -x^2 - 2x - 1
  18. -x^2 - 5x + 24
  19. -x^2 + 18x - 72
  20. -x^2 + 25x - 150
  21. x^2 + 21x + 108
  22. -x^2 + 11x - 30
  23. x^2 + 12x - 64
  24. x^2 - 17x - 60

Review Answers

  1. (x + 1) (x + 9)
  2. (x + 5) (x + 10)
  3. (x + 7) (x + 3)
  4. (x + 12) (x + 4)
  5. (x - 3) (x - 8)
  6. (x - 7) (x - 6)
  7. (x - 11) (x - 3)
  8. (x - 5) (x - 4)
  9. (x - 2) (x + 7)
  10. (x - 3) (x + 9)
  11. (x - 6) (x + 13)
  12. (x - 4) (x + 8)
  13. (x - 15) (x + 3)
  14. (x - 10) (x + 5)
  15. (x - 8) (x + 5)
  16. (x - 8) (x + 7)
  17. - (x + 1) (x + 1)
  18. - (x - 3) (x + 8)
  19. - (x - 6) (x - 12)
  20. - (x - 15) (x - 10)
  21. (x + 9) (x + 12)
  22. - (x - 5) (x - 6)
  23. (x - 4) (x + 16)
  24. (x - 20) (x + 3)

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CK.MAT.ENG.SE.1.Algebra-I.9.5

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