FACTORING CUBIC POLYNOMIALS

Every cubic polynomial will have 3 factors. To find those factors, we follow the following steps.

Step 1 :

We can find one linear factor of the given cubic polynomial using synthetic division.

Step 2 :

At the end of the first step, we will have quadratic factors. By factoring the quadratic equation, we can get other two factors.

Step 3 :

List all three factors.

Factor the following cubic polynomials :

Example 1 :

x3 - 2x2 - 5 x + 6

Solution :

Step 1 :

Let p(x)  =  x3 - 2x2 - 5 x + 6

Step 2 :

By dividing the cubic polynomial by 1, we get 0 as remainder. So (x - 1) is a factor. 

We can get the other two factors, by factoring the quadratic polynomial x- x - 6.

x- x - 6 = (x - 3)(x + 2)

Step 3 :

The required factors are (x - 1) (x - 3) and (x + 2).

Example 2 :

4x3 - 7x + 3

Solution :

Step 1 :

Let p(x)  =  4x3 - 7x + 3

Step 2 :

(x-1) is one of the factors.

We get the other two factors, by factoring the quadratic polynomial 4x+ 4x -3.

  =  4x+ 6x - 2x - 3 (decomposing the middle term)

  = 2x(2x + 3) - 1(2x + 3)

= (2x - 1)(2x + 3)

Step 3 :

So, the factors are (x - 1)(2x - 1)(2x + 3).

Example 3 :

x- 23x+ 142x - 120

Solution :

Step 1 :

Let p(x) = x- 23x+ 142x - 120

Step 2 :

(x - 1) is a factor.

x- 22x + 120 = (x - 10)(x - 12)

Step 3 :

So, the factors are (x - 1)(x - 10)(x - 12).

Example 4 :

4x- 5x+ 7x - 6

Solution :

Step 1 :

Let p(x) = 4x- 5x+ 7x - 6

Step 2 :

(x - 1) is one of the factors.

4x- x + 6 is not factorable.

Step 3 :

So, the factors are (x-1)(4x2-x+6).

Example 5 :

 x3 - 7x + 6

Solution :

Step 1 :

Let p(x) x3 - 7x + 6

Step 2 :

x+ x - 6 = (x + 3)(x - 2)

Step 3 :

So, the factors are (x - 1)(x + 3)(x - 2).

Example 6 :

 x3 + 13x+ 32x + 20

Solution :

Step 1 :

Let p(x) = x3 + 13x+ 32x + 20

Step 2 :

(x + 1) is one of the factors.

x+ 12x + 20 = (x + 10)(x + 2)

Step 3 :

So, the factors are (x + 1)(x + 10)(x + 2).

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