SOLVING CUBIC EQUATIONS

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

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

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

x² - 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 :

x³ - 23x² + 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 :

4x³ - 5x² + 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

4x² - x + 6 = 0

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

x = [-b ± √(b² - 4ac)]/2a

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

x = (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

x² + 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 + 13x² + 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

x² + 12x + 20 = 0

Factor and solve.

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

x = -2 or x = -10

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

Question 9 :

Find the remainder of 

f(x) = x³ + x² - 6x - 7

divided by x + 2

Solution :

x + 2 = 0

x = -2

f(-2) = (-2)³ + (-2)² - 6(-2) - 7

= -8 + 4 + 12 - 7

= -15 + 16

= 1

While dividing the polynomial by x + 2, we get the remainder 1.

Question 10 :

Find the value of k if 

f(x) = 3(x² + 3x - 4) - 8(x - k) is divisible by x.

Solution :

Since it is divsible by x, the x = 0

f(0) = 3(0² + 3(0) - 4) - 8(0 - k)

Since it is divisible by x, the remainder will be 0.

0 = 3(0 + 0 - 4) - 8(-k)

0 = -12 + 8k

8k = 12

k = 12/8

k = 3/2

So, the value of k is 3/2.

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