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

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

where \begin{align*}a, b,\end{align*} and \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 \begin{align*}a, b,\end{align*} and \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 \begin{align*}c = 0\end{align*}. For example, for the quadratic \begin{align*}ax^2 + bx\end{align*}, the common factor is \begin{align*}x\end{align*} and this expression is factored as \begin{align*}x(ax + b)\end{align*}. Now we’ll see how to factor quadratics where \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 \begin{align*}a = 1, b\end{align*} is positive and \begin{align*}c\end{align*} is positive. The quadratic trinomials will take the form

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

You know from multiplying binomials that when you multiply two factors \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:

\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:

\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.

\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 \begin{align*}m\end{align*} and \begin{align*}n\end{align*} where

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

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

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

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

Example 1

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

Solution

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

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

We want two numbers \begin{align*}m\end{align*} and \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:

\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 \begin{align*}(x + 2)(x + 3)\end{align*}.

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

\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 \begin{align*}x^2 + 7x + 12\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 12 can be written as the product of the following numbers:

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

Example 3

Factor \begin{align*}x^2 + 8x + 12\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 12 can be written as the product of the following numbers:

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

Example 4

Factor \begin{align*}x^2 + 12x + 36\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 36 can be written as the product of the following numbers:

\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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