Aarav Rangel

2022-02-10

How do you multiply ${\left(9-2{x}^{4}\right)}^{2}$?

liofila3w7

Explanation:
${\left(9-2{x}^{4}\right)}^{2}$
$=\left(9-2{x}^{4}\right)\left(9-2{x}^{4}\right)$
$=81-18{x}^{4}-18{x}^{4}+4{x}^{8}$
$=81-36{x}^{4}+4{x}^{8}$

Pierce Carey

Multiply/Simplify/Expand:
${\left(9-2{x}^{4}\right)}^{2}$
Use the square of a difference:
${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$,
where:
a=9, and $b=2{x}^{4}$
Plug in the known values.
${\left(9-2{x}^{4}\right)}^{2}={9}^{2}-\left(2\cdot 9\cdot 2{x}^{4}\right)+{\left(2{x}^{4}\right)}^{2}$
Simplify ${9}^{2}$ to 81.
${\left(9-2{x}^{4}\right)}^{2}=81-\left(2\cdot 9\cdot 2{x}^{4}\right)+{\left(2{x}^{4}\right)}^{2}$
Apply the multiplicative distributive property: ${\left(ab\right)}^{m}={a}^{m}{b}^{m}$
${\left(9-2{x}^{4}\right)}^{2}=81-\left(2\cdot 9\cdot 2{x}^{4}\right)+{2}^{2}\cdot {\left({x}^{4}\right)}^{2}$
Apply power rule: ${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$
${\left(9-2{x}^{4}\right)}^{2}=81-\left(2\cdot 9\cdot 2{x}^{4}\right)+{2}^{2}\cdot {x}^{8}$
Simplify.
${\left(9-2{x}^{4}\right)}^{2}=81-36{x}^{4}+4{x}^{8}$
Rearrange the equation in descending order.
${\left(9-2{x}^{4}\right)}^{2}=4{x}^{8}-36{x}^{4}+81$

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