How to factor polynomials

Question

1.) 125x^3-8

2.) x^2-9x-22

3.) 2x^2-5x+2

4.) x^3-x^2+5x-5

5.) 2x^3-3x^2+4x-6

 

 

Answer 1

It looks like this is a review of several different methods for factoring polynomials.

 

 

 

(1) DIFFERENCE OF CUBES

 

 

Sometimes, you'll have to recognize a special form, like a difference of squares (x² - y²) or a sum or difference of cubes (x³ + y³ or x³ - y³). Try to memorize the formulas for how each one is factored:

 

a² - b² = (a + b)(a - b)

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

 

 

In this case, you've got a difference of cubes. (5x)³ is 125x³ and (2)³ is 8. So, factor using the difference of cubes formula, where a = 5x and b = 2:

 

125x³ - 8 = (5x - 2)(25x² + (5x)(2) + 2²) = (5x - 2)(25x² + 10x + 4)

 

 

You'll want to make sure you're finished by trying to factor all the new factors. One good place to look is anywhere there's still an x². 25x² + 10x + 4 can't be factored nicely, though, so we'll leave it as-is. (Use the Quadratic Formula to see that its roots are imaginary.)

 

 

 

(2) BASIC QUADRATIC FACTORING

 

This is the method you'll use most often, whenever there's nothing in front of the x² term (a = 1). Try to think of a pair of numbers that multiply together to be -22 (c) and add together to be -9 (b).

 

1×-22 = -22

-1×22 = -22

2×-11 = -22

-2×11 = -22

 

The only pair that add up to be -9 is 2 and -11. So:

 

x² - 9x - 22 = (x + 2)(x - 11)

 

 

 

(3) LESS BASIC QUADRATIC FACTORING

 

Now that a ≠ 0, it gets a bit harder to come up with factors. You might need to use the Quadratic Formula to find factors (or to find that there are no rational factors).

 

Try thinking of a pair of numbers that multiply together to be 4 (a×c = 2×2) and add together to be -5.

 

1×4 = 4

-1×-4 = 4

2×2 = 4

-2×-2 = 4

 

The only pair that add up to be -5 is -1 and -4. So, replace the middle term with those numbers:

 

2x² - 5x + 2

2x² - x - 4x + 2

 

Now, factor by grouping (which we'll go over in more depth for number 4):

 

(2x² - x) + (-4x + 2)

x(2x - 1) - 2(2x - 1)

(x - 2)(2x - 1)

 

 

 

(4) FACTORING BY GROUPING

 

When we factor by grouping, we'll group two pairs of terms together and start by factoring out each pair's greatest common factor (whatever they have in common):

 

(x³ - x²) + (5x - 5)

x²(x - 1) + 5(x - 1)

 

You know you've done this right if the expressions in parentheses are equal. Otherwise, you'll need to make sure you factored everything out or try factoring out a -1. The final step is essentially to factor (x - 1) out of both of those terms, leaving you with:

 

(x² + 5)(x - 1)

 

As usual, check to make sure you can't factor anything further.

 

 

 

(5) FACTORING BY GROUPING

 

We finish with another fairly standard factor-by-grouping problem:

 

(2x³ - 3x²) + (4x - 6)

x²(2x - 3) + 2(2x - 3)

(x² + 2)(2x - 3)