How do you simplify $7 {(x^{4})}^{5} {(- 4 x^{2})}^{2}$ $7(x^4)^5 (-4x^2)^2$ ?

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Answer 1 · 2016-04-08T09:57:20

$112 x^{24}$ $112x^24$

As ${(x^{m})}^{n} = x^{m n}$ $(x^m)^n=x^(mn)$ ,

$7 {(x^{4})}^{5} \cdot {(- 4 x^{2})}^{2}$ $7(x^4)^5*(-4x^2)^2$

= $7 x^{4 \cdot 5} \cdot {(- 4)}^{2} \cdot x^{2 \cdot 2}$ $7x^(4*5)*(-4)^2*x^(2*2)$

= $7 x^{20} \cdot 16 \cdot x^{4}$ $7x^20*16*x^4$

= $112 \cdot x^{20 + 4}$ $112*x^(20+4)$

= $112 x^{24}$ $112x^24$

How do you simplify 7(x^4)^5 (-4x^2)^27(x4)5(−4x2)2 7(x^4)^5 (-4x^2)^2?