Question 14 Marks
Divide$: 36a^3x^5 − 24a^4x^4 + 18a^5x^3$ by $− 6a^3x^3.$
Answer
View full question & answer→$36 a^3 x^5-24 a^4 x^4+18 a^5 x^3 \text { by }-6 a^3 x^3$
$=\frac{36 a^3 x^5-24 a^4 x^4+18 a^5 x^3}{-6 a^3 x^3}$
$=\frac{36 a^3 x^5}{-6 a^3 x^3}-\frac{24 a^4 x^4}{-6 a^3 x^3}+\frac{18 a^5 x^3}{-6 a^3 x^3}$
$=-6 a^{3-3} \cdot x^{5-3}+4 a^{4-3} \cdot x^{4-3}-3 a^{5-3} \cdot x^{3-3}$
$=-6 a^0 x^2+4 a^1 x^2-3 a^2 x^0$
$=-6 x^2+4 a c-3 a^2 \ldots \ldots \ldots \ldots . .\left(\because x^0 \text { or } y^0=1\right)$
$=\frac{36 a^3 x^5-24 a^4 x^4+18 a^5 x^3}{-6 a^3 x^3}$
$=\frac{36 a^3 x^5}{-6 a^3 x^3}-\frac{24 a^4 x^4}{-6 a^3 x^3}+\frac{18 a^5 x^3}{-6 a^3 x^3}$
$=-6 a^{3-3} \cdot x^{5-3}+4 a^{4-3} \cdot x^{4-3}-3 a^{5-3} \cdot x^{3-3}$
$=-6 a^0 x^2+4 a^1 x^2-3 a^2 x^0$
$=-6 x^2+4 a c-3 a^2 \ldots \ldots \ldots \ldots . .\left(\because x^0 \text { or } y^0=1\right)$