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Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Write quadratic equations in standard form.
• Factor quadratic expressions for different coefficient values.

## Write Quadratic Expressions in Standard Form

Quadratic polynomials are polynomials of the 2nd\begin{align*}2^{nd}\end{align*} degree. The standard form of a quadratic polynomial is written as

ax2+bx+c\begin{align*}ax^2 + bx + c\end{align*}

where a,b,\begin{align*}a, b,\end{align*} and c\begin{align*}c\end{align*} stand for constant numbers. Factoring these polynomials depends on the values of these constants. In this section we’ll learn how to factor quadratic polynomials for different values of a,b,\begin{align*}a, b,\end{align*} and c\begin{align*}c\end{align*}. (When none of the coefficients are zero, these expressions are also called quadratic trinomials, since they are polynomials with three terms.)

You’ve already learned how to factor quadratic polynomials where c=0\begin{align*}c = 0\end{align*}. For example, for the quadratic ax2+bx\begin{align*}ax^2 + bx\end{align*}, the common factor is x\begin{align*}x\end{align*} and this expression is factored as x(ax+b)\begin{align*}x(ax + b)\end{align*}. Now we’ll see how to factor quadratics where c\begin{align*}c\end{align*} is nonzero.

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

First, let’s consider the case where a=1,b\begin{align*}a = 1, b\end{align*} is positive and c\begin{align*}c\end{align*} is positive. The quadratic trinomials will take the form

x2+bx+c\begin{align*}x^2 + bx +c\end{align*}

You know from multiplying binomials that when you multiply two factors (x+m)(x+n)\begin{align*}(x + m)(x + n)\end{align*}, you get a quadratic polynomial. Let’s look at this process in more detail. First we use distribution:

(x+m)(x+n)=x2+nx+mx+mn\begin{align*}(x + m)(x + n) = x^2 + nx + mx + mn\end{align*}

Then we simplify by combining the like terms in the middle. We get:

(x+m)(x+n)=x2+(n+m)x+mn\begin{align*}(x + m)(x + n) = x^2 + (n + m) x +mn\end{align*}

So to factor a quadratic, we just need to do this process in reverse.

We see that x2+(n+m)x+mnis the same form asx2+bx+c\begin{align*}& \text{We see that} \qquad \qquad \qquad \quad \ x^2 + (n + m)x + mn \\ & \text{is the same form as} \qquad \qquad x^ 2 + bx + c\end{align*}

This means that we need to find two numbers m\begin{align*}m\end{align*} and n\begin{align*}n\end{align*} where

n+m=bandmn=c\begin{align*} n + m = b \qquad \qquad \text{and} \qquad \qquad mn = c\end{align*}

The factors of x2+bx+c\begin{align*}x^2 + bx + c\end{align*} are always two binomials

(x+m)(x+n)\begin{align*}(x + m)(x + n)\end{align*}

such that n+m=b\begin{align*}n + m = b\end{align*} and mn=c\begin{align*}mn = c\end{align*}.

Example 1

Factor x2+5x+6\begin{align*}x^2 + 5x + 6\end{align*}.

Solution

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

(x)(x)\begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

We want two numbers m\begin{align*}m\end{align*} and n\begin{align*}n\end{align*} that multiply to 6 and add up to 5. A good strategy is to list the possible ways we can multiply two numbers to get 6 and then see which of these numbers add up to 5:

6=166=23andand1+6=72+3=5This is the correct choice.\begin{align*}& 6 = 1 \cdot 6 && \text{and} && 1 + 6 = 7\\ & 6 = 2 \cdot 3 && \text{and} && 2 + 3 = 5 \qquad This \ is \ the \ correct \ choice.\end{align*}

So the answer is (x+2)(x+3)\begin{align*}(x + 2)(x + 3)\end{align*}.

We can check to see if this is correct by multiplying (x+2)(x+3)\begin{align*}(x + 2)(x + 3)\end{align*}:

x+2x+3 3x+6x2+2xx2+5x+6\begin{align*}& \quad \quad \quad x + 2\\ & \underline{\;\;\;\;\;\;\;\;\;\;x + 3}\\ & \quad \quad \ 3x + 6\\ & \underline{x^2 + 2x\;\;\;\;\;\;}\\ & x^2 + 5x + 6\end{align*}

Example 2

Factor x2+7x+12\begin{align*}x^2 + 7x + 12\end{align*}.

Solution

We are looking for an answer that is a product of two binomials in parentheses: (x)(x)\begin{align*}(x\;\;\;\;\;)(x\;\;\;\;)\end{align*}

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

12=11212=2612=34andandand1+12=132+6=83+4=7This is the correct choice.\begin{align*}& 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 This \ is \ the \ correct \ choice.\end{align*}

The answer is (x+3)(x+4)\begin{align*}(x + 3)(x + 4)\end{align*}.

Example 3

Factor x2+8x+12\begin{align*}x^2 + 8x + 12\end{align*}.

Solution

We are looking for an answer that is a product of two binomials in parentheses: (x)(x)\begin{align*}(x\;\;\;\;\;)(x\;\;\;\;)\end{align*}

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

12=11212=2612=34andandand1+12=132+6=8This is the correct choice.3+4=7\begin{align*}& 12 = 1 \cdot 12 && \text{and} && 1 + 12 = 13\\ & 12 = 2 \cdot 6 && \text{and} && 2 + 6 = 8 \qquad This \ is \ the \ correct \ choice.\\ & 12 = 3 \cdot 4 && \text{and} && 3 + 4 = 7\end{align*}

The answer is (x+2)(x+6)\begin{align*}(x + 2)(x + 6)\end{align*}.

Example 4

Factor x2+12x+36\begin{align*}x^2 + 12x + 36\end{align*}.

Solution

We are looking for an answer that is a product of two binomials in parentheses: (x)(x)\begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

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

36=13636=21836=31236=4936=66andandandandand1+36=372+18=203+12=154+9=136+6=12This is the correct choice.\begin{align*}& 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 This \ is \ the \ correct \ choice.\end{align*}

The answer is \begin{align*}(x + 6)(x + 6)\end{align*}.

## 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 5

Factor \begin{align*}x^2 - 6x + 8\end{align*}.

Solution

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

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:

\begin{align*}8 = 1 \cdot 8 \quad \quad \text{and} \quad \quad 1 + 8 = 9\end{align*}

but also

\begin{align*}8 = (-1) \cdot (-8) \quad \quad \text{and} \quad \quad -1 + (-8) = -9\end{align*}

and

\begin{align*}8 = 2 \cdot 4 \quad \quad \text{and} \quad \quad 2 + 4 = 6\end{align*}

but also

\begin{align*}8 = (-2) \cdot (-4) \quad \quad \text{and} \quad \quad -2 + (-4) = -6 \qquad This \ is \ the \ correct \ choice.\end{align*}

The answer is \begin{align*}(x - 2)(x - 4)\end{align*}. We can check to see if this is correct by multiplying \begin{align*}(x - 2)(x - 4)\end{align*}:

\begin{align*}& \quad \quad \quad x - 2\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;x - 4}\\ & \quad \ - \ 4x + 8\\ & \underline{x^2 - \ 2x\;\;\;\;\;\;\;}\\ & x^2 - \ 6x + 8\end{align*} The answer checks out.

Example 6

Factor \begin{align*}x^2 - 17x + 16\end{align*}.

Solution

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

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

\begin{align*}& 16 = 1 \cdot 16 && \text{and} && 1 + 16 = 17\\ & 16 = (-1) \cdot (-16) && \text{and} && -1 + (-16) = -17 \qquad 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\end{align*} The answer is \begin{align*}(x - 1)(x - 16)\end{align*}.

In general, whenever \begin{align*}b\end{align*} is negative and \begin{align*}a\end{align*} and \begin{align*}c\end{align*} 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 7

Factor \begin{align*}x^2 + 2x - 15\end{align*}.

Solution

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;\;)\end{align*}

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

\begin{align*}& -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 This \ is \ the \ correct \ choice.\\ & -15 = 3 \cdot (-5) && \text{and} && 3 + (-5) = -2\end{align*}

The answer is \begin{align*}(x - 3)(x +5)\end{align*}.

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

\begin{align*}& \quad \quad \ \ x - \ 3\\ & \underline{\;\;\;\;\;\;\;\;\; x + \;5\;}\\ & \quad \quad 5x - 15\\ & \underline{x^2 - 3x\;\;\;\;\;\;\;\;}\\ & x^2 + 2x - 15\end{align*} The answer checks out.

Example 8

Factor \begin{align*}x^2 - 10x - 24\end{align*}.

Solution

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

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

\begin{align*}& -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 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\end{align*}

The answer is \begin{align*}(x - 12) (x + 2)\end{align*}.

Example 9

Factor \begin{align*}x^2 + 34x - 35\end{align*}.

Solution

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

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

\begin{align*}& -35 = -1 \cdot 35 && \text{and} && -1 + 35 = 34 \qquad 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\end{align*}

The answer is \begin{align*}(x - 1)(x + 35)\end{align*}.

## Factor when a = - 1

When \begin{align*}a = -1\end{align*}, 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 10

Factor \begin{align*}-x^2 + x + 6\end{align*}.

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:

\begin{align*}-x^2 + x + 6 = -(x^2 - x - 6)\end{align*}

We’re looking for a product of two binomials in parentheses: \begin{align*}-(x\;\;\;\;)(x\;\;\;\;)\end{align*}

Now our job is to factor \begin{align*}x^2 - x - 6\end{align*}.

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

\begin{align*}& -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 This \ is \ the \ correct \ choice.\end{align*}

The answer is \begin{align*}-(x - 3)(x + 2)\end{align*}.

## Lesson Summary

• A quadratic of the form \begin{align*}x^2 + bx + c\end{align*} factors as a product of two binomials in parentheses: \begin{align*}(x + m)(x + n)\end{align*}
• If \begin{align*}b\end{align*} and \begin{align*}c\end{align*} are positive, then both \begin{align*}m\end{align*} and \begin{align*}n\end{align*} are positive.

Example: \begin{align*}x^2 + 8x + 12\end{align*} factors as \begin{align*}(x + 6)(x + 2)\end{align*}.

• If \begin{align*}b\end{align*} is negative and \begin{align*}c\end{align*} is positive, then both \begin{align*}m\end{align*} and \begin{align*}n\end{align*} are negative.

Example: \begin{align*}x^2 - 6x + 8\end{align*} factors as \begin{align*}(x - 2)(x - 4)\end{align*}.

• If \begin{align*}c\end{align*} is negative, then either \begin{align*}m\end{align*} is positive and \begin{align*}n\end{align*} is negative or vice-versa.

Example: \begin{align*}x^2 + 2x -15\end{align*} factors as \begin{align*}(x + 5)(x - 3)\end{align*}.

Example: \begin{align*}x^2 + 34x - 35\end{align*} factors as \begin{align*}(x + 35)(x - 1)\end{align*}.

• If \begin{align*}a = -1\end{align*}, factor out -1 from each term in the trinomial and then factor as usual. The answer will have the form: \begin{align*}-(x + m)(x + n)\end{align*}

Example: \begin{align*}-x^2 + x + 6\end{align*} factors as \begin{align*}-(x - 3)(x +2)\end{align*}.

## Review Questions

1. \begin{align*}x^2 + 10x + 9\end{align*}
2. \begin{align*}x^2 + 15x + 50\end{align*}
3. \begin{align*}x^2 + 10x + 21\end{align*}
4. \begin{align*}x^2 + 16x + 48\end{align*}
5. \begin{align*}x^2 - 11x + 24\end{align*}
6. \begin{align*}x^2 - 13x + 42\end{align*}
7. \begin{align*}x^2 - 14x + 33\end{align*}
8. \begin{align*}x^2 - 9x + 20\end{align*}
9. \begin{align*}x^2 + 5x - 14\end{align*}
10. \begin{align*}x^2 + 6x - 27\end{align*}
11. \begin{align*}x^2 + 7x - 78\end{align*}
12. \begin{align*}x^2 + 4x - 32\end{align*}
13. \begin{align*}x^2 - 12x - 45\end{align*}
14. \begin{align*}x^2 - 5x - 50\end{align*}
15. \begin{align*}x^2 - 3x - 40\end{align*}
16. \begin{align*}x^2 - x - 56\end{align*}
17. \begin{align*}-x^2 - 2x - 1\end{align*}
18. \begin{align*}-x^2 - 5x + 24\end{align*}
19. \begin{align*}-x^2 + 18x - 72\end{align*}
20. \begin{align*}-x^2 + 25x - 150\end{align*}
21. \begin{align*}x^2 + 21x + 108\end{align*}
22. \begin{align*}-x^2 + 11x - 30\end{align*}
23. \begin{align*}x^2 + 12x - 64\end{align*}
24. \begin{align*}x^2 - 17x - 60\end{align*}
25. \begin{align*}x^2 + 5x - 36\end{align*}

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