Using the factor theorem, please find the prime factors of the polynomial

Question

x^3 + 5x^2 + 2x - 8

Answer 1

The polynomial is x ³ + 5x ² + 2x – 8

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

If p/q is a rational zero, then p  is a factor of 8 and q  is a factor of 1.

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

The possible values for q  are ± 1.

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

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

p/q	1	5	2	-8
-1	1	4	-2	-6
1	1	6	8	0

Since f (1) = 0, x = 1 is a zero. The depressed polynomial is  x ² + 6x + 8.

Here f (1) = 0, therefore by factor theorem (x - 1) is a factor.

Therefore x^3 + 5x^2 + 2x - 8

=> (x - 1) (x ² + 2x + 4x+ 8)

=> (x - 1) { x (x + 2) + 4(x + 2) }

=> (x - 1) { (x + 2) (x + 4) }

=> (x - 1) (x + 2) (x + 4)

Therefore (x - 1) , (x + 2) and (x + 4) are the factors of the polynomial

 

Answer 2

Factor Theorem : Let f(x) be a polynomial such that f(c) = 0 for some constant c. Then x - c is a factor of f(x). Conversely, if x - c is a factor of f(x), then f(x) = 0.

The Factor Theorem is an algebraic topic that involves finding the roots (or zeros) of a polynomial function. There are two methods that one can use in discovering the roots:

• Trial and Error Method
• Berry Method (for binomial functions only)

Trial and Error Method:

With the trial and error method you will need to try and guess a number that makes the polynomial equal to zero. This sounds like a time consuming task but most algebra texts will have a zero of a function between negative 5 and positive 5. Say you try plugging in 4 into the above equation. If you do, you will get an answer of f(x) = 6 which is not equal to zero so 4 cannot be a root. But say you plug in 1, now you do get f(x) = 0 so 1 is a root so you can conclude that (x-1) is indeed a factor of the polynomial.

Let the function is f(x) = x ³ + 5x ² + 2x – 8.

Let us check x = 1 by trial and error method. Then, f(1) = (1)³ + 5(1)² + 2(1) - 8 = 1 +  5 + 2 - 8 = 0.

According to factor theorem, if f(1) = 0, then (x - 1) is a factor of the polynomial f(x).

Therefore, (x - 1) is a factor of the given polynomial.

By dividing the given polynomial with (x - 1).

x - 1) x ³ + 5x ² + 2x – 8 (x² + 6x + 8

 (-)   x ³ -  x ²

       _________________

               6x ² + 2x

        (-)   6x ² - 6x

       _________________

                         8x - 8

       (-)              8x - 8

       _________________

                           0.

By dividing the given polynomial with (x - 1), we get x² + 6x + 8.

Factoring trinomial, x² + 6x + 8 as follows.

=> (x ² + 2x + 4x+ 8)

=> x (x + 2) + 4(x + 2)

=> (x + 2) (x + 4)

=> (x + 2) (x + 4).

The prime factors of x ³ + 5x ² + 2x – 8 are (x - 1), (x + 2) and (x + 4).