(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.

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.