How do you simplify $7 x^{3} y^{8} (- 4 x^{2} y^{4})$ $7x^3y^8(-4x^2y^4)$ ?

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How do you simplify $7 x^{3} y^{8} (- 4 x^{2} y^{4})$ $7x^3y^8(-4x^2y^4)$ ?

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Answer 1 · 2017-03-19T17:43:51

$- 28 x^{5} y^{12}$ $-28x^5y^12$

Explanation:

The simplification involves simple multiplication of constant numbers and application of addition rule when multiplying exponents with same base. Accordingly,

$7 x^{3} y^{8} (- 4 x^{2} y^{4}) = - 28 x^{3 + 2} y^{8 + 4} = - 28 x^{5} y^{12}$ $7x^3 y^8 (-4x^2 y^4)= -28 x^(3+2) y^(8+4)= -28x^5y^12$

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How do you simplify $7 x^{3} y^{8} (- 4 x^{2} y^{4})$ $7x^3y^8(-4x^2y^4)$ ?

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How do you simplify $7 x^{3} y^{8} (- 4 x^{2} y^{4})$ $7x^3y^8(-4x^2y^4)$ ?