"How do you simplify ((4s3t−2)(2t−3u5)8u7)2" was answered by our community

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

ku3sump8

Answered question

2022-02-03

How do you simplify

{(\frac{(4 s^{3} t^{- 2}) (2 t - 3 u^{5})}{8 u^{7}})}^{2}

Answer & Explanation

Belen Hebert

Beginner2022-02-04Added 8 answers

Step 1
${(\frac{(4 s^{3} t^{- 2}) (2 t - 3 u^{5})}{8 u^{7}})}^{2}$
$= (\frac{4^{2} {(s^{3})}^{2} {(\frac{1}{t^{2}})}^{2} {(2 t - 3 u^{5})}^{2}}{8^{2} {(u^{7})}^{2}})$
$= \frac{16 s^{3 \times 2} (\frac{1^{2}}{t^{2 \times 2}}) {(2 t - 3 u^{5})}^{2}}{64 u^{7 \times 2}}$
$= \frac{⧸ 16 s^{6} (\frac{1}{t^{4}}) {(2 t - 3 u^{5})}^{2}}{⧸ 16 \times 4 u^{14}}$
$= \frac{s^{6} {(2 t - 3 u^{5})}^{2}}{4 t^{4} u^{14}}$

Darien Myers

Beginner2022-02-05Added 12 answers

Step 1
${(\frac{(4 s^{3} t^{- 2}) (2 t - 3 u^{5})}{8 u^{7}})}^{2}$
Simplify inside the bracket first:
$= {(\frac{⧸ 4 s^{3} (2 t - 3 u^{5})}{⧸ 8^{2} t^{2} u^{7}})}^{2}$
Square each factor in the bracket:
$\frac{s^{6} {(2 t - - 3 u^{5})}^{2}}{2^{2} t^{4} u^{14}}$
We could also square the binomial but there seems little point.
We would not be able to cancel any factors (there will be + and -signs.
There will not be like terms nor a common factor
The answer is already in factor form.

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