How do you factor $a^{6} + 125$ $a^6+125$ ?

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How do you factor $a^{6} + 125$ $a^6+125$ ?

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Answer 1 · 2015-11-20T14:37:08

$a^{6} + 125 = (a^{2} + 5) (a^{4} - 5 a^{2} + 25)$ $a^6+125 = (a^2+5)(a^4-5a^2+25)$

$= (a^{2} + 5) (a^{2} + \sqrt{15} a + 5) (a^{2} - \sqrt{15} a + 5)$ $=(a^2+5)(a^2+sqrt(15)a+5)(a^2-sqrt(15)a+5)$

Explanation:

First use the sum of cubes identity:

$A^{3} + B^{3} = (A + B) (A^{2} - A B + B^{2})$ $A^3+B^3=(A+B)(A^2-AB+B^2)$

with $A = a^{2}$ $A=a^2$ and $B = 5$ $B=5$ to find:

$a^{6} + 125 = {(a^{2})}^{3} + 5^{3} = (a^{2} + 5) ({(a^{2})}^{2} - (a^{2}) (5) + 5^{2})$ $a^6+125 = (a^2)^3+5^3 = (a^2 + 5)((a^2)^2 - (a^2)(5) + 5^2)$

$= (a^{2} + 5) (a^{4} - 5 a^{2} + 25)$ $= (a^2+5)(a^4-5a^2+25)$

Neither $a^{2} + 5$ $a^2+5$ nor $a^{4} - 5 a^{2} + 25$ $a^4-5a^2+25$ have linear factors with Real coefficients, but $a^{4} - 5 a^{2} + 25$ $a^4-5a^2+25$ will have quadratic factors with Real coefficients.

Consider:

$(a^{2} + k a + 5) (a^{2} - k a + 5) = a^{4} + (10 - k^{2}) a^{2} + 25$ $(a^2+ka+5)(a^2-ka+5) = a^4+(10-k^2)a^2+25$

If we let $k = \sqrt{15}$ $k = sqrt(15)$ then:

$(a^{2} + k a + 5) (a^{2} - k a + 5) = a^{4} + (10 - 15) a^{2} + 25 = a^{4} - 5 a^{2} + 25$ $(a^2+ka+5)(a^2-ka+5) = a^4+(10-15)a^2+25 = a^4-5a^2+25$

So:

$a^{4} - 5 a^{2} + 25 = (a^{2} + \sqrt{15} a + 5) (a^{2} - \sqrt{15} a + 5)$ $a^4-5a^2+25 = (a^2+sqrt(15)a+5)(a^2-sqrt(15)a+5)$

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How do you factor $a^{6} + 125$ $a^6+125$ ?

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