Question

Divide, Write the quotient and the remainder.
$40a^3÷(-10a)$

Answer 1

$40a^3÷(-10a)$

$=\frac{40a^3}{-10a}$

$=\frac{40a^3}{-10a^1}$

$=-4a^{3-1}$

$=-4a^2$

$\therefore 40a^3÷(-10a)=-4a^2$

Since, Dividend $=$ Divisor $\times$ Quotient $+$ Remainder

$\Rightarrow 40a^3=(-10a)\times (-4a^2)+0$

i.e., Quotient $=-4a^2$

Remainder $=0$