FACTORING CUBIC POLYNOMIALS

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 x² - x - 6.

x² - 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 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).

Example 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).

Example 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).

Example 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).

Example 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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