Aarav Rangel

2022-02-10

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

liofila3w7

Beginner2022-02-11Added 13 answers

Explanation:

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

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

$=81-18{x}^{4}-18{x}^{4}+4{x}^{8}$

$=81-36{x}^{4}+4{x}^{8}$

Pierce Carey

Beginner2022-02-12Added 10 answers

Multiply/Simplify/Expand:

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

Use the square of a difference:

$(a-b)}^{2}={a}^{2}-2ab+{b}^{2$ ,

where:

a=9, and$b=2{x}^{4}$

Plug in the known values.

$(9-2{x}^{4})}^{2}={9}^{2}-(2\cdot 9\cdot 2{x}^{4})+{\left(2{x}^{4}\right)}^{2$

Simplify$9}^{2$ to 81.

$(9-2{x}^{4})}^{2}=81-(2\cdot 9\cdot 2{x}^{4})+{\left(2{x}^{4}\right)}^{2$

Apply the multiplicative distributive property:$\left(ab\right)}^{m}={a}^{m}{b}^{m$

$(9-2{x}^{4})}^{2}=81-(2\cdot 9\cdot 2{x}^{4})+{2}^{2}\cdot {\left({x}^{4}\right)}^{2$

Apply power rule:$\left({a}^{m}\right)}^{n}={a}^{m\cdot n$

$(9-2{x}^{4})}^{2}=81-(2\cdot 9\cdot 2{x}^{4})+{2}^{2}\cdot {x}^{8$

Simplify.

$(9-2{x}^{4})}^{2}=81-36{x}^{4}+4{x}^{8$

Rearrange the equation in descending order.

${(9-2{x}^{4})}^{2}=4{x}^{8}-36{x}^{4}+81$

Use the square of a difference:

where:

a=9, and

Plug in the known values.

Simplify

Apply the multiplicative distributive property:

Apply power rule:

Simplify.

Rearrange the equation in descending order.