Question

Divide $42(a^4-13a^3+36a^2)$ by $7a(a-4)$

Answer 1

$42(a^4-13a^3+36a^2)=2\times 3\times 7\times a\times a\times (a^2-13a+36)$
$=2\times 3\times 7\times a\times a\times (a^2-9a-4a+36)$
$=2\times 3\times 7\times a\times a\times \left[a\right(a-9)-4(a-9\left)\right]$
$=2\times 3\times 7\times a\times a\times \left[\right(a-9\left)\right(a-4\left)\right]$
$=2\times 3\times 7\times a\times a\times (a-9)(a-4)$
$42(a^4-13a^3+36a^2)÷7a(a-4)=2\times 3\times 7\times a\times a\times (a-9)(a-4)÷7a(a-4)$
$=6a(a-9)$