## SOLVING CUBIC EQUATIONS

Every cubic equation would have 3 solutions. To solve a cubic equation, either the method factoring or synthetic division can be used.

If a cubic equation equation is factorable, factor it by grouping and solve for the variable.

Synthetic division can be used for the cubic equations which can not be factored by grouping.

Solve each of the following cubic equations.

Question 1 :

x3 + 2x2 - x - 2 = 0

Solution :

The given cubic equation can be factored by grouping.

Factor and solve.

x3 + 2x2 - x - 2 = 0

x2(x + 2) - 1(x + 2) = 0

(x + 2)(x2 - 1) = 0

 x + 2 = 0x = -2 x2 - 1 = 0x2 - 12 = 0(x + 1)(x - 1) = 0x = -1 or x = 1

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

Question 2 :

2x3 - x2 - 4x + 2 = 0

Solution :

The given cubic equation can be factored by grouping.

Factor and solve.

2x3 - x2 - 4x + 2 = 0

x2(2x - 1) - 2(2x - 1) = 0

(2x - 1)(x2 - 2) = 0

 2x - 1 = 0x = 1/2 x2 - 2 = 0x2 = 2x = ±√2

So, the solution is {1/2, √2, -√2}.

Question 3 :

x3 - 2x2 - 5x + 6 = 0

Solution :

Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.

1 is one of the roots. The other roots can be determined by solving the quadratic equation

x- x - 6 = 0

Factor and solve.

(x - 3)(x + 2) = 0

x = 3 or x = -2

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

Question 4 :

4x3 - 7x + 3 = 0

Solution :

Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.

1 is one of the roots. The other roots can be determined by solving the quadratic equation

4x2 + 4x - 3 = 0

Factor and solve.

4x2 + 6x - 2x - 3 = 0

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

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

x = -3/2 or x = 1/2

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

Question 5 :

x- 23x+ 142x - 120 = 0

Solution :

Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.

1 is one of the roots. The other roots can be determined by solving the quadratic equation

x2 - 22x + 120 = 0

Factor and solve.

(x  - 10)(x - 12) = 0

x = 10 or x = 12

So, the solution is {1, 10, 12}.

Question 6 :

4x- 5x+ 7x - 6 = 0

Solution :

Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.

1 is one of the roots. The other roots can be determined by solving the quadratic equation

4x- x + 6 = 0

This quadratic equation can not be solved by factoring. So use quadratic formula and solve.

= [-b ± √(b2 - 4ac)]/2a

a = 4, b = -1 and c = 6,

= (1 ± √-95)/8

For the given cubic equation, there is only one real root, that is 1. The remaining two roots are imaginary.

Question 7 :

x3 - 7x + 6 = 0

Solution :

Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.

1 is one of the roots. The other roots can be determined by solving the quadratic equation

x+ x - 6 = 0

Factor and solve.

(x + 3)(x - 2) = 0

x = -3 or x = 2

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

Question 8 :

x3 + 13x+ 32x + 20 = 0

Solution :

Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.

-1 is one of the roots. The other roots can be determined by solving the quadratic equation

x+ 12x + 20 = 0

Factor and solve.

(x + 2)(x + 10) = 0

x = -2 or x = -10

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

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