## SOLVING CUBIC POLYNOMIAL EQUATIONS

Every cubic polynomial will have 3 solutions. To find those solutions, we follow the steps given below.

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 :

Equating each factor to zero, we can find the solution.

Solve each of the following polynomial

Question 1 :

x3 - 2x2 - 5 x + 6  =  0

Solution :

Step 1 :

Let p(x)  =  x3 - 2x2 - 5 x + 6  =  0

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)

x3-2x2-5x+6  =  0

(x-1) (x-3) (x+2)  =  0

 x-1  =  0x  =  1 x-3  =  0x  =  3 x+2  =  0x  =  -2

So, the solution is {1, 2, 3}.

Question 2 :

4x3 - 7x + 3  =  0

Solution :

Step 1 :

Let p(x)  =  4x3 - 7x + 3  =  0

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).

Step 4 :

Equating each factor to zero, we get

 x-1  =  0x  =  1 2x-1  =  0x  =  1/2 2x+3  =  0x  =  -3/2

So, the solution is {-3/2, 1, 1/2}.

Question 3 :

x3-23x2+142x-120  =  0

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)

Step 4 :

Equating each factor to zero, we get

 x-1  =  0x  =  1 x-10  =  0x  =  10 x-12  =  0x  =  12

So, the solution is {1, 10, 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).

Step 4 :

Comparing the quadratic equation 4x2-x+6 with the general form of quadratic equation, we get

a  =  4, b  =  -1 and c  =  6.

=  (-b±√b2 -4ac)/2a

x  =  (1±√1 -96)/8

x  =  (1±√-95)/8 and x  =  1

So, the solution is {1, (1±√-95)/8}.

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).

Step 4 :

Equating factors to zero, we get

 x-1  =  0x  =  1 x+3  =  0x  =  -3 x-2  =  0x  =  2

So, the solution is {-3, 1, 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).

By equating each factor to zero, we get

 x+1  =  0x  =  -1 x+10  =  0x  =  -10 x+2  =  0x  =  -2

So, the solution is {-1, -10, -2}.

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