SOLVING CUBIC EQUATIONS WORKSHEET

Solve each of the following cubic equations.

1) x3 + 2x2 - x - 2 = 0

2) 2x3 - x2 - 4x + 2 = 0

3) x3 - 2x2 - 5x + 6 = 0

4) 4x3 - 7x + 3 = 0

5) x- 23x+ 142x - 120 = 0

6) 4x- 5x+ 7x - 6 = 0

7) x3 - 7x + 6 = 0

8)  x3 + 13x+ 32x + 20 = 0

1. Answer :

x3 + 2x2 - x - 2 = 0

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 = 0

x = -2

x2 - 1 = 0

x2 - 12 = 0

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

x = -1 or x = 1

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

2. Answer :

2x3 - x2 - 4x + 2 = 0

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 = 0

x = 1/2

x2 - 2 = 0

x2 = 2

x = ±√2

So, the solution is {1/2, √2, -√2}.

3. Answer :

x3 - 2x2 - 5x + 6 = 0

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}.

4. Answer :

4x3 - 7x + 3 = 0

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}.

5. Answer :

x- 23x+ 142x - 120 = 0

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}.

6. Answer :

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

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. 

7. Answer :

x3 - 7x + 6 = 0

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}.

8. Answer :

 x3 + 13x+ 32x + 20 = 0

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}.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Operations with Negative Numbers

    Jan 16, 22 10:35 PM

    Operations with Negative Numbers

    Read More

  2. Playing with Numbers

    Jan 16, 22 08:14 PM

    Playing with Numbers

    Read More

  3. Word Names for Numbers

    Jan 16, 22 12:37 PM

    Word Names for Numbers

    Read More