How do you factor $4 x^{5} - 9 x^{3}$ $4x^5 - 9x^3$ ?

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How do you factor $4 x^{5} - 9 x^{3}$ $4x^5 - 9x^3$ ?

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Answer 1 · 2017-04-24T07:48:19

$= x^{3} (2 x + 3) (2 x - 3)$ $= x^3(2x+3)(2x-3)$

Explanation:

Always take out a common factor first if there is one.

$4 x^{5} - 9 x^{3}$ $4x^5-9x^3$

$= x^{3} (4 x^{2} - 9) \leftarrow$ $= x^3(4x^2 -9)" "larr$ difference of squares

$= x^{3} (2 x + 3) (2 x - 3)$ $= x^3(2x+3)(2x-3)$

To recognise Difference of squares:

2 terms
minus sign
perfect squares
even indices

$x^{2} = y^{2} = (x + y) (x - y)$ $x^2 =y^2 = (x+y)(x-y)$

Answer 2 · 2017-04-24T08:00:58

$x^{3} (2 x + 3) (2 x - 3)$ $x^3(2x+3)(2x-3)$

Explanation:

$4 x^{5} - 9 x^{3}$ $4x^5-9x^3$

Factor out $x^{3}$ $x^3$

$x^{3} (4 x^{2} - 9)$ $x^3(4x^2-9)$

Here we realize that both $4 x^{2}$ $4x^2$ and $9$ $9$ are perfect squares. So, we factor them as.

$(\sqrt{A} + \sqrt{B}) (\sqrt{A} - \sqrt{B})$ $color(red)((sqrtA+sqrtB) (sqrtA-sqrtB))$
I this case $A = 4 x^{2}$ $color(red)A=4x^2$ and $B = 9$ $color(red)B=9$

So,
$x^{3} (\sqrt{4 x^{2}} + \sqrt{9}) (\sqrt{4 x^{2}} - \sqrt{9})$ $x^3(sqrt(4x^2)+sqrt9)(sqrt(4x^2)-sqrt9)$

Complete the square roots to get your final answer

$x^{3} (2 x + 3) (2 x - 3)$ $x^3(2x+3)(2x-3)$

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