How do you factor $f^3+2f^2-64f-128$ ?

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Answer 1 · 2017-12-03T10:10:08

Required factors are : $(f+8)(f-8)(f+2)$

The problem is to factor

$f^3 + 2f^2 - 64f - 128$

Observe the groupings collected for further simplification:

$color(green)((f^3 + 2f^2)$ $color(blue)(- 64f - 128)$ $color(red)(Expression.1)$

The group $color(green)((f^3 + 2f^2)$ can be factored as

$color(green)(f^2(f + 2))$ $" "color(red)(Res.1)$

The group $color(blue)(- 64f - 128)$ can be factored as

$color(blue)(-64(f+2))$ $" "color(red)(Res.2)$

Using our intermediate results $color(red)(Res.1)$ and $color(red)(Res.2)$ we can write our $color(red)(Expression.1)$ as

$color(green)(f^2(f+2)-64(f+2))$

We can now factor them as

$(f^2 - 64)(f+2)$ $color(red)(Expression.2)$

Next, we rewrite $(f^2 - 64)$ as

$(f^2 - 8^2)$

Using the factoring rule "Difference of Squares" we get

$(f+8)(f - 8)$

Using the above result and our $color(red)(Expression.2)$
we get all of our required factors:

$color(blue)((f+8)(f-8)(f+2))$

How do you factor f^3+2f^2-64f-128f^3+2f^2-64f-128?