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

To factor 6x⁵ - 4x³ - 16x, we can first factor out the greatest common factor (GCF). Each term is divisible by 2x so we can divide each term by a common factor of 2x:

6x⁵ - 4x³ - 16x

2x(3x⁴ - 2x² - 8)

Notice, the powers are x⁴, x², and a constant. This means we can use substitution and treat it like a quadratic expression. We can let:

u = x²

Then:

3x⁴ - 2x² - 8 becomes 3u² - 2u - 8

3u² - 2u - 8 is now in the quadratic form ax² + bx + c so we can use the ac method to factor this. In our expression, a = 3, b = -2, c = -8. To use the ac method, we first multiply a by c:

3 x (-8) = -24

Now, find two numbers that multiply to -24 and add up to -2. Those numbers are:

4 and -6

So, we split the middle term:

3u² + 4u - 6u - 8 We can use grouping here to group terms

(3u² + 4u) + (-6u - 8) Take out factor from each group

u(3u + 4) - 2(3u + 4)

(u - 2)(3u + 4)

We can now substitute the x² back in to u:

(u - 2)(3u + 4) = (x² - 2)(3x² + 4)

So, now the expression is:

2x(x² - 2)(3x² + 4)

We can further factor x² - 2 since that is a difference of squares which has the form:

a² - b² = (a + b)(a - b)

So x² - 2 can be written as (x)² - (√2)² so we can now infer that a is x and b is √2 and it can be written as:

(x)² - (√2)² = (x + √2)(x - √2)

Altogether, it becomes:

2x(x + √2)(x - √2)(3x² + 4)

The 3x² + 4 cannot be factored further over real numbers since it does not have real solutions. We can check this by equaling it to 0:

3x² + 4 = 0

3x² = -4

x² = -4 / 3

Notice, x² = negative number and there is no real number whose square is negative which is why it cannot be factored over the real numbers. Therefore, the final factored form as you also stated is:

2x(x + √2)(x - √2)(3x² + 4)

Answer 3

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 4

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)