Factor completely over the real numbers 6x5-4x3-16x

Question

Factor completely over the real numbers (the solution is 2x(3x2+4)(x+√2)(x−√2). Please show a step-by step process for solving this problem.

1). 6x⁵-4x³-16x

Answer 1

Given 6x⁵-4x³-16x,

1. Factor out the GCF, i.e. 2x ⇒ 2x(3x⁴-2x²-8)
2. Since 3x⁴-2x²-8 is quadratic in form, use the factors of the product of the leading coefficient and the constant term that add up to the coefficient of the middle term in order to rewrite the middle term as follows: 2x(3x⁴-6x²+4x²-8)
3. Next, factor by grouping: 2x[ (3x⁴-6x²) + (4x²-8) ] = 2x[3x²(x²-2)+4(x²-2)] = 2x(3x²+4)(x²-2)
4. Since x²-4 = (x)²-(√2)² = (x+√2)(x-√2). Factoring the difference of squares.
5. So that, 6x⁵-4x³-16x = 2x(3x²+4)(x+√2)(x-√2)

Note: One may also use the rational zero theorem, synthetic division, and the factor theorem to factor the given 5th degree polynomial completely over the real numbers.

Good luck!

Answer 2

6x^5 -4x^3 -16x

= 2x(3x^4 -2x^2 -8)

let y = x^2

= 2x(3y^2 -2x -8) = 2x(3y +4)(y -2)

= 2x(3x^2 +4)(x^2-2)

most people stop there. but if you want to do irrational factors next step

= 2x(3x^2 +4)(x-sqr2)(x+sqr2)

a 5th degree does have 5 roots = 5 factors, so here they are:

2

x

3x^2+4

x+sqr2

x-sqr2

are the Big Five, as good as any

Answer 3

We can factor out the greatest common factor from each term:

6x⁵-4x³-8x = 2x(3x⁴-2x²-8)

Factor the expression inside the parenthesis.

(3x⁴-2x²-8) = (3x²+4)(x²-2)

The term x²-2 can be further factored:

(x²-2) = (x+√2)(x-√2)

Therefore,

6x⁵-4x³-8x = (3x²+4)(x+√2)(x-√2)