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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