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

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Answer 1 · 2018-06-14T19:09:14

$2 a (a^{2} + 2 a + 4)$ $2a(a^2+2a+4)$

We immediately recognize that al terms have an $a$ $a$ in common, so we can factor that out. We get

$a (2 a^{2} + 4 a + 8)$ $acolor(blue)((2a^2+4a+8))$

We also notice that all of the terms I have in blue have a $2$ $2$ in common that we can factor out. Doing this, we now have

$2 a (a^{2} + 2 a + 4)$ $2acolor(blue)((a^2+2a+4))$

What I have in blue, let's think for a moment:

Are there any two numbers that sum up to $2$ $2$ and have a product of $4$ $4$ ?

You may have got stuck. There are no such numbers. This means this is the most we can factor this expression with real numbers. Our answer is

$2 a (a^{2} + 2 a + 4)$ $2a(a^2+2a+4)$

Hope this helps!

Answer 2 · 2018-06-14T19:12:16

See a solution process below:

Factor a $2 a$ $color(red)(2a)$ form each term in the expression:

$2 a^{3} + 4 a^{2} + 8 a \Rightarrow$ $2a^3 + 4a^2 + 8a =>$

$(2 a \cdot a^{2}) + (2 a \cdot 2 a) + (2 a \cdot 4) \Rightarrow$ $(color(red)(2a) * a^2) + (color(red)(2a) * 2a) + (color(red)(2a) * 4) =>$

$2 a (a^{2} + 2 a + 4)$ $color(red)(2a)(a^2 + 2a + 4)$

The term within the parenthesis cannot be factored further.

How do you factor 2a3+4a2+8a 2a^3+4a^2+8a?