How do you multiply $(- 3x ^ { 2} y ^ { 3} z ^ { 4} ) ^ { 2} ( 2y ) ^ { 2}$ ?

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Answer 1 · 2017-07-01T04:11:42

$36x^4y^8z^8$

Indices rule

$a^1*a^2=a^(2+1)=a^3$

$(a^1)^2=a^(1*2)=a^2$

Therefore,

$(−3x^2y^3z^4)^2*(2y)^2$

$=((-3)^2*x^(2*2)y^(3*2)z^(4*2))*(2^2y^2)$

$=(9x^4y^6z^8)(4y^2)$

$=36x^4y^(6+2)z^8$

$=36x^4y^8z^8$

How do you multiply (- 3x ^ { 2} y ^ { 3} z ^ { 4} ) ^ { 2} ( 2y ) ^ { 2}(- 3x ^ { 2} y ^ { 3} z ^ { 4} ) ^ { 2} ( 2y ) ^ { 2}?