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Factorization of Quadratic Expressions

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What if you had a quadratic expression like x^2 + 9x + 14 in which all the coefficients were positive? How could you factor that expression? After completing this Concept, you'll be able to factor quadratic expressions like this one with positive coefficient values.

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CK-12 Foundation: 0908S Factoring Quadratic Expressions

Guidance

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

ax^2 + bx + c

where a, b, and c 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, and c . (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 . For example, for the quadratic ax^2 + bx , the common factor is x and this expression is factored as x(ax + b) . Now we’ll see how to factor quadratics where c is nonzero.

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

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

x^2 + bx +c

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

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

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

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

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

& \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

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

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

The factors of x^2 + bx + c are always two binomials

(x + m)(x + n)

such that n + m = b and mn = c .

Example A

Factor x^2 + 5x + 6 .

Solution

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

(x\;\;\;\;)(x\;\;\;\;)

We want two numbers m and n 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 = 1 \cdot 6 && \text{and} && 1 + 6 = 7\\& 6 = 2 \cdot 3 && \text{and} && 2 + 3 = 5 \qquad 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 x + 2\\& \underline{\;\;\;\;\;\;\;\;\;\;x + 3}\\& \quad \quad \ 3x + 6\\& \underline{x^2 + 2x\;\;\;\;\;\;}\\& x^2 + 5x + 6

The answer checks out.

Example B

Factor x^2 + 7x + 12 .

Solution

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

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 This \ is \ the \ correct \ choice.

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

Example C

Factor x^2 + 8x + 12 .

Solution

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

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 This \ is \ the \ correct \ choice.\\& 12 = 3 \cdot 4 && \text{and} && 3 + 4 = 7

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

Watch this video for help with the Examples above.

CK-12 Foundation: Factoring Quadratic Expressions

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.

Guided Practice

Factor x^2 + 12x + 36 .

Solution

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

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 This \ is \ the \ correct \ choice.

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

Practice

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+14x+45
  6. x^2+15x+50
  7. x^2+22x+40
  8. x^2+15x+56
  9. x^2+2x+1
  10. x^2+10x+24
  11. x^2+17x+72
  12. x^2+25x+150

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