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# Factorization of Quadratic Expressions with Negative Coefficients

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What if you had a quadratic expression like $x^2 - 3x - 10$ or $-x^2 - 4x -4$ in which some or all the coefficients were negative? How could you factor that expression? After completing this Concept, you'll be able to factor quadratic expressions like these for various negative coefficient values.

### Guidance

In the previous concept, we saw how to factor quadratic expressions whose coefficients were all positive. In this concept we will now see what happens when we factor quadratic expressions where some of the coefficients are negative.

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

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

#### Example A

Factor $x^2 - 6x + 8$ .

Solution

We are looking for an answer that is a product of two binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;)$

When negative coefficients are involved, we have to remember that negative factors may be involved also. The number 8 can be written as the product of the following numbers:

$8 = 1 \cdot 8 \quad \quad \text{and} \quad \quad 1 + 8 = 9$

but also

$8 = (-1) \cdot (-8) \quad \quad \text{and} \quad \quad -1 + (-8) = -9$

and

$8 = 2 \cdot 4 \quad \quad \text{and} \quad \quad 2 + 4 = 6$

but also

$8 = (-2) \cdot (-4) \quad \quad \text{and} \quad \quad -2 + (-4) = -6.$

The last option is the correct choice. The answer is $(x - 2)(x - 4)$ . We can check to see if this is correct by multiplying $(x - 2)(x - 4)$ :

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

#### Example B

Factor $x^2 - 17x + 16$ .

Solution

We are looking for an answer that is a product of two binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;)$

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

$& 16 = 1 \cdot 16 && \text{and} && 1 + 16 = 17\\& 16 = (-1) \cdot (-16) && \text{and} && -1 + (-16) = -17 \qquad (Correct \ choice)\\& 16 = 2 \cdot 8 && \text{and} && 2 + 8 = 10\\& 16 = (-2) \cdot (-8) && \text{and} && -2 + (-8) = -10\\& 16 = 4 \cdot 4 && \text{and} && 4 + 4 = 8\\& 16 = (-4) \cdot (-4) && \text{and} && -4 + (-4) = -8$

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

In general, whenever $b$ is negative and $a$ and $c$ are positive, the two binomial factors will have minus signs instead of plus signs.

Factor when a = 1 and c is Negative

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

#### Example C

Factor $x^2 + 2x - 15$ .

Solution

We are looking for an answer that is a product of two binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;\;)$

Once again, we must take the negative sign into account. The number -15 can be written as the product of the following numbers:

$& -15 = -1 \cdot 15 && \text{and} && -1 + 15 = 14\\& -15 = 1 \cdot (-15) && \text{and} && 1 + (-15) = -14\\& -15 = -3 \cdot 5 && \text{and} && -3 + 5 = 2 \qquad \qquad (Correct \ choice)\\& -15 = 3 \cdot (-5) && \text{and} && 3 + (-5) = -2$

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

We can check to see if this is correct by multiplying:

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

#### Example D

Factor $x^2 - 10x - 24$ .

Solution

We are looking for an answer that is a product of two binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;)$

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

$& -24 = -1 \cdot 24 && \text{and} && -1 + 24 = 23\\& -24 = 1 \cdot (-24) && \text{and} && 1 + (-24) = -23\\& -24 = -2 \cdot 12 && \text{and} && -2 + 12 = 10\\& -24 = 2 \cdot (-12) && \text{and} && 2 + (-12) = -10 \qquad (Correct \ choice)\\& -24 = -3 \cdot 8 && \text{and} && -3 + 8 = 5\\& -24 = 3 \cdot (-8) && \text{and} && 3 + (-8) = -5\\& -24 = -4 \cdot 6 && \text{and} && -4 + 6 = 2\\& -24 = 4 \cdot (-6) && \text{and} && 4 + (-6) = -2$

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

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 and then apply the methods you learned so far in this section

#### Example E

Factor $-x^2 + x + 6$ .

Solution

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

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

We’re looking for a product of two binomials in parentheses: $-(x\;\;\;\;)(x\;\;\;\;)$

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

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

$& -6 = -1 \cdot 6 && \text{and} && -1 + 6 = 5\\& -6 = 1 \cdot (-6) && \text{and} && 1 + (-6) = -5\\& -6 = -2 \cdot 3 && \text{and} && -2 + 3 = 1\\& -6 = 2 \cdot (-3) && \text{and} && 2 + (-3) = -1 \qquad (Correct \ choice)$

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

Watch this video for help with the Examples above.

### Vocabulary

• A quadratic of the form $x^2 + bx + c$ factors as a product of two binomials in parentheses: $(x + m)(x + n)$
• If $b$ and $c$ are positive, then both $m$ and $n$ are positive.
• If $b$ is negative and $c$ is positive, then both $m$ and $n$ are negative.
• If $c$ is negative, then either $m$ is positive and $n$ is negative or vice-versa.
• If there is a negative in front of $x^2$ , factor out -1 from each term in the trinomial and then factor as usual. The answer will have the form: $-(x + m)(x + n)$

### Guided Practice

Factor $x^2 + 34x - 35$ .

Solution

We are looking for an answer that is a product of two binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;)$

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

$& -35 = -1 \cdot 35 && \text{and} && -1 + 35 = 34 \qquad (Correct \ choice)\\& -35 = 1 \cdot (-35) && \text{and} && 1 + (-35) = -34\\& -35 = -5 \cdot 7 && \text{and} && -5 + 7 = 2\\& -35 = 5 \cdot (-7) && \text{and} && 5 + (-7) = -2$

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

### Practice

1. $x^2 - 11x + 24$
2. $x^2 - 13x + 42$
3. $x^2 - 14x + 33$
4. $x^2 - 9x + 20$
5. $x^2 + 5x - 14$
6. $x^2 + 6x - 27$
7. $x^2 + 7x - 78$
8. $x^2 + 4x - 32$
9. $x^2 - 12x - 45$
10. $x^2 - 5x - 50$
11. $x^2 - 3x - 40$
12. $x^2 - x - 56$
13. $-x^2 - 2x - 1$
14. $-x^2 - 5x + 24$
15. $-x^2 + 18x - 72$
16. $-x^2 + 25x - 150$
17. $x^2 + 21x + 108$
18. $-x^2 + 11x - 30$
19. $x^2 + 12x - 64$
20. $x^2 - 17x - 60$
21. $x^2 + 5x - 36$