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Factor Quadratics: ax2 + bx + c

Introduction

Quadratics are polynomials (specifically trinomials) that can be represented by the expression ax2 + bx + c, where “a”, “b”, and “c” are numbers.  Factoring is easiest when “a” is equal to 1.  For example:  The two numbers in the factors generally add up to “b” and multiply to “c”.  For example:

 Factor x2 + 6x + 8

 This trinomial factors into two binomials  of the form (x + m)(x + n), where “m” and “n” are whole numbers.  Start by writing:

 

          (x     )(x      )

The two numbers in the factors generally add up to “b” and multiply to “c”.  The factors here should add up to 6 and multiply to 8.  Not many pairs of numbers multiply to 8, but we do know that 4 × 2 = 8.  

(x + 4)(x + 2)

These numbers add up to 6, so they work in this factoring.  Our final answer:

x2 + 6x + 8 = (x + 4)(x + 2)

 

Check our answer

 

Factoring can be checked by doing the process backward and multiplying polynomials.

 

Factoring_harder_trinomials_vis_1

 

The answer checks. 

 

 

A more difficult quadratic like 2x2 + 7x + 6 can be factored using similar techniques.  Click on the lesson tab above to see how to factor it.


Lesson

When factoring the quadratic ax2 + bx + c, first consider the possible factors of “a”.

Factor 2x2 + 7x + 6

 

The “a” term is 2, and the only possible factors of 2 are 2 × 1.  Begin by writing:

(2x      )(x        )

The “c” term is 6, so the two numbers in our equation must multiply to 6.  The possibilities here are 3 × 2 and 2 × 3.  Simply use trial-and-error for easier problems like this one to figure out which combination works:

 

Factoring_harder_trinomials_vis_2

Final answer: 2x2 + 7x + 6 = (2x + 3)(x + 2)

 

 

The sample problem above shows that the order of the numbers is important.  (2x + 2)(x + 3) does not work, but (2x + 3)(x + 2) does work.

 

The Box Method

The box method can be used for factoring just as it can be used for multiplying polynomials.

     Factor 3x2 + 13x + 14

 

Factoring_harder_trinomials_vis_3

 

One more step is required to get to the final answer.  In this final step, find the two remaining numbers.  The number at the top right must equal 7x when multiplied by 7.  The number at the bottom left must equal 6x when multiplied by 3x.

 

Factoring_harder_trinomials_vis_4

 

Now that the box has been factored, the answer can (and should) be written as the product of two factors.

Answer: 3x2 + 13x + 14 = (3x + 7)(x + 2)      

 

Remember to take the product of ac in order to find the (bottom left) and (top right) numbers in the box when using this method.  Here is another example.

       Factor 5x2 + 4x – 12

 

Factoring_harder_trinomials_vis_5

 Answer: 5x2 + 4x – 12 = (x + 2)(5x – 6) 

     

 

Example 1Factor the quadratics:

 

Factoring_harder_trinomials_vis_6

 

 

Solution:

 

Factoring_harder_trinomials_vis_7

 

 

In some cases, a number can be factored out of the quadratic to simplify the problem before you get started. 

     Factor 4x2 + 22x + 10

 

Factoring_harder_trinomials_vis_8

Final answer: 4x2 + 22x + 10 =2(2x2 + 11x + 10) = 2(2x + 1)(x + 5)

*Note: It is very easy to forget about the 2 that was factored out at the beginning.  This 2 must be included in the final answer.

 

  

Example 2:  Factor the quadratics:

 

Factoring_harder_trinomials_vis_9

 

 

Solution:

 

Factoring_harder_trinomials_vis_10

 

 

So the first thing to do when encountering a quadratic is to see whether a number can be factored out of the three terms.  Then use the steps given in this lesson to complete the factoring.

 

Try It

Factor each quadratic.  Write “not factorable” if it cannot be factored.

 

1)  5a2 + 10a + 25 

2)  3b2 – 4b – 15  

3)  6c2 + 39c + 63 

4)  10d2 – 29d – 21 

5)  e2 – 15e + 28 

6)  12a2 + 4a - 16 

*Hint:  It is easy to peek at the answers below, so it is recommended that you try all problems on a separate piece of paper, then check answers all at once.

 

 

 

Scroll down for answers:

 

 

 

 

 

 

 

 

 

 

Answers:

1)  5a2 + 10a + 25  =  5(a2 + 10a + 25) = 5(a + 5)(a + 5)

 

2)  3b2 – 4b – 15  =  (3b + 5)(b – 3)

 

3)  6c2 + 39c + 63  =  3(2c2 + 13c + 21)  =  3(2c + 7)(c + 3)

4)  10d2 – 29d – 21  =  (5d + 3)(2d – 7)

5)  e2 – 15e + 28  =  (2e – 7)(e – 4)

 

6)  12a2 + 4a - 16  =  4(3a2 + a – 4)  =  4(3a + 4)(a – 1)

 

 

Related Links:

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