How do you factor the expression $12 t^{8} - 75 t^{4}$ $12t^8-75t^4$ ?

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How do you factor the expression $12 t^{8} - 75 t^{4}$ $12t^8-75t^4$ ?

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Answer 1 · 2016-10-17T06:24:50

$12 t^{8} - 75 t^{4} = 3 t^{4} (2 t^{2} - 5) (2 t^{2} + 5)$ $12t^8-75t^4 = 3t^4(2t^2-5)(2t^2+5)$

$12 t^{8} - 75 t^{4} = 3 t^{4} (\sqrt{2} t - \sqrt{5}) (\sqrt{2} t + \sqrt{5}) (2 t^{2} + 5)$ $color(white)(12t^8-75t^4) = 3t^4(sqrt(2)t-sqrt(5))(sqrt(2)t+sqrt(5))(2t^2+5)$

Explanation:

$color(white)()$
Difference of squares

The difference of squares identity can be written:

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

We will use this a couple of times.

$color(white)()$
Factor $bb(12t^8 - 75t^4)$

First note that both terms are divisible by $3t^4$ , so separate that out as a factor first:

$12t^8-75t^4 = 3t^4(4t^4-25)$

$color(white)(12t^8-75t^4) = 3t^4((2t^2)^2-5^2)$

$color(white)(12t^8-75t^4) = 3t^4(2t^2-5)(2t^2+5)$

$color(white)(12t^8-75t^4) = 3t^4((sqrt(2)t)^2-(sqrt(5))^2)(2t^2+5)$

$color(white)(12t^8-75t^4) = 3t^4(sqrt(2)t-sqrt(5))(sqrt(2)t+sqrt(5))(2t^2+5)$

The remaining quadratic $(2t^2+5)$ does not factor further with Real coefficients, since $2t^2+5 >= 5 > 0$ for any Real value of $t$ .

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How do you factor the expression $12 t^{8} - 75 t^{4}$ $12t^8-75t^4$ ?

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