PRACTICE PROBLEMS ON CUBIC EQUATION USING SYNTHETIC DIVISION

Completely factor and find all zeros for each polynomial :

(1)  x³ + 4x² + 5x + 2

(2)  x⁴- x³ + 14x² - 16x - 32

(3)  5x³+29x²+19x-5

(4)  4x³ - 9x² + 6x - 1

(5)  3x⁴- 10x³ - 24x² - 6x + 5

(6)  3x³ + 9x² + 4x + 12

Problem 1 :

x³ + 4x² + 5x + 2

Solution :

Let p(x)  =  x³ + 4x² + 5x + 2

By dividing the cubic polynomial by 1, we get 12 ≠ 0.

Dividing by -1, we get 0 as remainder. So, -1 is one of the solutions.

From the bottom row, we create a quadratic polynomial. By factoring this, we will get two factors.

x²+3x+2  = (x+1) (x+2)

x³+4x²+5x+2  =  0

(x+1) (x+1) (x+2)  =  0

Equating each factor to zero, we get

x + 1  =  0, x + 1  =  0 and x + 2  =  0

x  =  -1, -1, -2

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

Problem 2 :

x⁴- x³ + 14x² - 16x - 32

Solution :

Let p(x)  =  x⁴- x³ + 14x² - 16x - 32

By dividing the fourth degree polynomial by 1, we get -34≠0.

By applying the value of x as -1 and 2, we get the remainder as 0.

So, the factors are (x+1) and (x–2)

So far we got two roots of the given polynomial.

By solving,

x² + 16  =  0

x²  =  -16

x  =  √-16

x  =  ±4i

So, it has a imaginary root.

x⁴- x³ + 14x² - 16x - 32

x + 1  =  0 and x - 2  =  0

x  =  -1, 2

So, the solution is {-1, 2, ±4i}

Problem 3 :

5x³+29x²+19x-5

Solution :

Let p(x)  =  5x³+29x²+19x–5

By dividing the cubic polynomial by 1, we get 38≠0. Dividing by -1, we get 0 as remainder. So, -1 is one of the solution.

By factoring quadratic polynomial 5x² + 24x – 5, we get

5x² + 24x - 5  =  (5x - 1) (x + 5)

(5x - 1) (x + 5)  =  0

Equating each factor to zero, we get

 x  =  -1, 1/5, -5

So, the solution is {-1, 1/5, -5}

Problem 4 :

4x³ - 9x² + 6x - 1

Solution :

Let p(x)  =  4x³ - 9x² + 6x - 1 

By dividing the cubic polynomial by 1, we get 0 as remainder. So, 1 is one of the zeroes of the polynomial.

By factoring quadratic polynomial 4x² - 5x + 1, we get

4x² - 5x + 1  =  (4x – 1) (x – 1)

The factors are (x – 1) (4x – 1) (x – 1)

Equating each factor to zero, we get

(x – 1) (4x – 1) (x – 1)  =  0

 x  =  1, 1/4, 1

So, the solution is {1, 1/4, 1}

Problem 5 :

3x⁴- 10x³ - 24x² - 6x + 5

Solution :

Let p(x)  =  3x⁴- 10x³ - 24x² - 6x + 5

By dividing the fourth degree polynomial by 1, we get -32≠0. So dividing by 5, we get 0 as remainder,

Both x  =  5 and x  =  -1 are the two zeros of given polynomial.

By factoring quadratic polynomial 3x² + 2x - 1, we get

3x² + 2x - 1  =  (3x – 1) (x + 1)

Equating each factor to zero, we get

(3x – 1) (x + 1)  =  0

x  = 1/3, -1

So, the solution is {5, -1, 1/3, -1}

Problem 6 :

3x³ + 9x² + 4x + 12

Solution :

Let p(x)  =  3x³ + 9x² + 4x + 12

By dividing the cubic polynomial by 1, we get 18≠0. Dividing by -3, we get 0 as remainder. So, -3 is one of the zeroes of the given polynomial.

By solving,

3x² + 4 =  0

3x²  =  -4

x  =  √-4/3

x  =  ±2/√3i

So, the solution is {-3, ±2/√3i}.

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