How do you factor $6 x^{5} - 51 x^{3} - 27 x$ $6x^5-51x^3-27x$ ?

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How do you factor $6 x^{5} - 51 x^{3} - 27 x$ $6x^5-51x^3-27x$ ?

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Answer 1 · 2016-02-05T21:22:47

$6 x^{5} - 51 x^{3} - 27 x$ $6x^5-51x^3-27x$

$= 3 x (2 x^{2} + 1) (x - 3) (x + 3)$ $= 3x(2x^2+1)(x-3)(x+3)$

$= 3 x (\sqrt{2} x - i) (\sqrt{2} x + i) (x - 3) (x + 3)$ $= 3x(sqrt(2)x-i)(sqrt(2)x+i)(x-3)(x+3)$

Explanation:

We will use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

with $a = x$ $a=x$ and $b = 3$ $b=3$ later...

First separate out the common factor $3 x$ $3x$ of the terms to find:

$6 x^{5} - 51 x^{3} - 27 x$ $6x^5-51x^3-27x$

$= 3 x (2 x^{4} - 17 x^{2} - 9)$ $= 3x(2x^4-17x^2-9)$

The remaining quartic factor is a quadratic in $x^{2}$ $x^2$ , which we can factor using the AC method with grouping. First find a pair of factors of $A C = 2 \cdot 9 = 18$ $AC=2*9=18$ which differ by $B = 17$ $B=17$ . The pair $18, 1$ $18, 1$ works. Then use this pair to split the middle term and factor by grouping:

$= 3 x (2 x^{4} - 18 x^{2} + x^{2} - 9)$ $= 3x(2x^4-18x^2+x^2-9)$

$= 3 x ((2 x^{4} - 18 x^{2}) + (x^{2} - 9))$ $= 3x((2x^4-18x^2)+(x^2-9))$

$= 3 x (2 x^{2} (x^{2} - 9) + 1 (x^{2} - 9))$ $= 3x(2x^2(x^2-9)+1(x^2-9))$

$= 3 x (2 x^{2} + 1) (x^{2} - 9)$ $= 3x(2x^2+1)(x^2-9)$

$= 3 x (2 x^{2} + 1) (x^{2} - 3^{2})$ $= 3x(2x^2+1)(x^2-3^2)$

$= 3 x (2 x^{2} + 1) (x - 3) (x + 3)$ $= 3x(2x^2+1)(x-3)(x+3)$

This is as far as we can go with Real coefficients, but if we allow Complex coefficients then this can be factored further as:

$= 3 x (\sqrt{2} x - i) (\sqrt{2} x + i) (x - 3) (x + 3)$ $= 3x(sqrt(2)x-i)(sqrt(2)x+i)(x-3)(x+3)$

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How do you factor $6 x^{5} - 51 x^{3} - 27 x$ $6x^5-51x^3-27x$ ?

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How do you factor 6x^5-51x^3-27x6x5−51x3−27x6x^5-51x^3-27x?

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How do you factor $6 x^{5} - 51 x^{3} - 27 x$ $6x^5-51x^3-27x$ ?