Simplify: $7 a^{3} (- 4 a^{4} + 6 a^{5})$

14 a^{7}

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14 a^{8}

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- 28 a^{7} + 42 a^{8}

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- 28 a^{7} + 42 a^{15}

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Solution

The correct option is C $- 28 a^{7} + 42 a^{8}$
$7 a^{3} (- 4 a^{4} + 6 a^{5})$
Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
$=$ $7 a^{3} . - 4 a^{4} + 7 a^{3} .6 a^{5}$
$=$ $- 28 a^{7} + 42 a^{8}$

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Similar questions

A = [\begin{matrix} 1 & 4 2 & 3 \end{matrix}] a n d A^{5} - 4 A^{4} - 7 A^{3} + 11 A^{2} - A - 10 I = A + x I,

6 a^{5} - 18 a^{3} + 42 a^{2}

Factorise:

$a^{6} - 7 a^{3} - 8$

Factorise
$7 a^{3} + 56 b^{3}$

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