How do you factor $6 y^{6} - 5 y^{3} - 4$ $6y^6-5y^3-4$ ?

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How do you factor $6 y^{6} - 5 y^{3} - 4$ $6y^6-5y^3-4$ ?

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Answer 1 · 2018-05-07T18:27:34

$(2 y^{3} + 1) (3 y^{3} - 4)$ $(2y^3+1)(3y^3-4)$

Explanation:

$using a substitution reduces the expression to a$ $"using a substitution reduces the expression to a"$
$usual quadratic$ $"usual quadratic"$

$let u = y^{3}$ $"let "u=y^3$

$\Rightarrow 6 y^{6} - 5 y^{3} - 4 = 6 u^{2} - 5 u - 4$ $rArr6y^6-5y^3-4=6u^2-5u-4$

$using the a-c method for factoring$ $"using the a-c method for factoring"$

$the factors of the product 6 \times - 4 = - 24$ $"the factors of the product "6xx-4=-24$

$which sum to - 5 are + 3 and - 8$ $"which sum to - 5 are + 3 and - 8"$

$split the middle term using these factors$ $"split the middle term using these factors"$

$6 u^{2} + 3 u - 8 u - 4 \leftarrow factor by grouping$ $6u^2+3u-8u-4larrcolor(blue)"factor by grouping"$

$= 3 u (2 u + 1) - 4 (2 u + 1)$ $=color(red)(3u)(2u+1)color(red)(-4)(2u+1)$

$take out the common factor (2 u + 1)$ $"take out the "color(blue)"common factor "(2u+1)$

$= (2 u + 1) (3 u - 4)$ $=(2u+1)(color(red)(3u-4))$

$change the substitution back into terms in y$ $"change the substitution back into terms in y"$

$\Rightarrow 6 y^{6} - 5 y^{3} - 4 = (2 y^{3} + 1) (3 y^{3} - 4)$ $rArr6y^6-5y^3-4=(2y^3+1)(3y^3-4)$

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How do you factor $6 y^{6} - 5 y^{3} - 4$ $6y^6-5y^3-4$ ?

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