How do you factor $a^{4} - 8 a^{3} + 16^{2}$ $a^4 - 8a^3 + 16^2$ ?

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How do you factor $a^{4} - 8 a^{3} + 16^{2}$ $a^4 - 8a^3 + 16^2$ ?

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Answer 1 · 2017-06-15T20:23:09

Assuming you mean $a^{4} - 8 a^{3} + 16 a^{2}$ $a^4 - 8a^3 + 16a^2$ , note the $a^{2}$ $a^2$ in common:

$a^{2} (a^{2} - 8 a + 16)$ $a^2(a^2 - 8a + 16)$

Then, we look for factors of $16$ $16$ that when added together in a certain way give $8$ $8$ . As it turns out, this is a perfect square.

$= a^{2} {(a - 4)}^{2}$ $= color(blue)(a^2(a - 4)^2)$

If you multiply this out using the "FOIL" method, you should get the same thing back that you started with:

$a^{2} {(a - 4)}^{2} = a^{2} (a^{2} - 4 a - 4 a + 16)$ $a^2(a - 4)^2 = a^2(a^2 - 4a - 4a + 16)$

$= a^{2} (a^{2} - 8 a + 16)$ $= a^2(a^2 - 8a + 16)$

$= a^{4} - 8 a^{3} + 16 a^{2}$ $= a^4 - 8a^3 + 16a^2$

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How do you factor $a^{4} - 8 a^{3} + 16^{2}$ $a^4 - 8a^3 + 16^2$ ?

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