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.

Factorize each of the following polynomial 

Question 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 x²-x-6.

x²-x-6  =  (x-3)(x+2)

Step 3 :

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

Question 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 4x²+4x-3.

  =  4x²+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).

Question 3 :

x³-23x²+142x-120

Solution :

Step 1 :

Let p(x)  =  x³-23x²+142x-120

Step 2 :

(x-1) is a factor.

x²-22x+120  =  (x-10)(x-12)

Step 3 :

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

Question 4 :

4x³-5x²+7x-6

Solution :

Step 1 :

Let p(x)  =  4x³-5x²+7x-6

Step 2 :

(x-1) is one of the factors.

4x²-x+6 is not factorable.

Step 3 :

So, the factors are (x-1)(4x²-x+6).

Question 5 :

 x³ - 7x + 6

Solution :

Step 1 :

Let p(x)  =   x³ - 7x + 6

Step 2 :

x²+x-6  =  (x+3)(x-2)

Step 3 :

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

Question 6 :

 x³ +13x²+32x+20

Solution :

Step 1 :

Let p(x)  =   x³ +13x²+32x+20

Step 2 :

(x+1) is one of the factors.

x²+12x+20  =  (x+10)(x+2)

Step 3 :

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

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