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** :

x^{3} - 2x^{2} - 5 x + 6

**Solution :**

**Step 1 :**

Let p(x) = x^{3} - 2x^{2} - 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^{2}-x-6.

x^{2}-x-6 = (x-3)(x+2)

**Step 3 :**

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

**Question 2** :

4x^{3} - 7x + 3

**Solution :**

**Step 1 :**

Let p(x) = 4x^{3} - 7x + 3

**Step 2 :**

(x-1) is one of the factors.

We get the other two factors, by factoring the quadratic polynomial 4x^{2}+4x-3.

= 4x^{2}+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** :

x^{3}-23x^{2}+142x-120

**Solution :**

**Step 1 :**

Let p(x) = x^{3}-23x^{2}+142x-120

**Step 2 :**

(x-1) is a factor.

x^{2}-22x+120 = (x-10)(x-12)

**Step 3 :**

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

**Question 4** :

4x^{3}-5x^{2}+7x-6

**Solution :**

**Step 1 :**

Let p(x) = 4x^{3}-5x^{2}+7x-6

**Step 2 :**

(x-1) is one of the factors.

4x^{2}-x+6 is not factorable.

**Step 3 :**

So, the factors are (x-1)(4x^{2}-x+6).

**Question 5** :

x^{3} - 7x + 6

**Solution :**

**Step 1 :**

Let p(x) = x^{3} - 7x + 6

**Step 2 :**

x^{2}+x-6 = (x+3)(x-2)

**Step 3 :**

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

**Question 6** :

x^{3} +13x^{2}+32x+20

**Solution :**

**Step 1 :**

Let p(x) = x^{3} +13x^{2}+32x+20

**Step 2 :**

(x+1) is one of the factors.

x^{2}+12x+20 = (x+10)(x+2)

**Step 3 :**

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

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