How to find all of the real and imaginary zeros of this polynomial function?

Question

1. f(x) = x^3-9x^2=26x-24 

2. h(x) = x^3-x^2-7x+15 

Answer 1

(2).

Identify Rational Zeros :

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation a_nx ⁿ + a_n – 1x ^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

The function is x ³- x ²- 7x + 15 = 0.

If p/q is a rational zero, then p is a factor of 15 and q is a factor of 1.

The possible values of p are ± 1, ± 3, ± 5, and ± 15.

The possible values for q are ± 1.

By the Rational Roots Theorem, the only possible rational roots are, p / q = ± 1, ± 3, ± 5, and ± 15.

Make a table for the synthetic division and test possible real zeros.

p/q	1	- 1	- 7	15
- 1	1	- 2	- 5	20
- 2	1	- 3	- 1	17
- 3	1	- 4	5	0

Since, h(-3) = 0, x = -3 is a zero. The depressed polynomial is  x ² - 4x + 5 = 0.

Completing the square Method :

The equation is x² - 4x + 5 = 0.

Separate variables and constants aside by subtracting 5 from each side.

x² - 4x + 5 - 5 = 0 - 5

x² - 4x = - 5

To change the expression (x² - 4x) into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

 Here x coefficient = - 4. So, (half the x coefficient)² = (- 4/2)²= 4.

Add 4 to each side.

x² - 4x + 4 = - 5 + 4

(x - 2)² = - 1

(x - 2)² = i².

x - 2 = ± i.

x = 2 ± i.

The value of x = - 3, x = 2 + i and x = 2 - i.

Answer 2

(1)

Given polynomial 

Step 1 :   Identify possible rational roots

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation

 a_nx ⁿ + a_n – 1x ^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

The function is x ³- 9x ²+ 26x - 24 = 0.

If p/q is a rational zero, then p is a factor of 1 and q is a factor of 24.

The possible values of p are  ± 1 , ± 2 , ± 3 , ± 4.

The possible values for q are ± 1.

By the Rational Roots Theorem, the only possible rational roots are, p / q = ± 1,  ± 2, ± 3 and ± 4..

Step 2 :   Synthetic division

Make a table for the synthetic division and test possible real zeros.

Make a table for the synthetic division and test possible real zeros.

p/q	1	- 9	26	-24
1	1	-8	18	-6
2	1	-7	12	0

Since, f(2) = 0, x = 2 is a zero.

The depressed polynomial is  x ² - 7x + 12 = 0.

Step 3 :   Factorisation

By factor by grouping.

x ² - 3x - 4x + 24 = 0

x(x - 3) - 4(x - 3) = 0

Factor : (x - 3)(x - 4) = 0

Apply zero product property.

x - 3 = 0 and x - 4 = 0

⇒ x = 3 and x = 4.

Solution :

The roots of the function are x = 1, x = - 2, x = 3.