Question

Multiply $-4x{y}^3$ and $6x^{2}y$ verify your result for $x=2$ and $y=1$.

Answer 1

$(-4x{y}^3)\times \left(6x^{2}y\right)$

$=(-4\times 6)(x\times x^2)(y^3\times y)$

$=-24x^3{y}^4$ [Since, $a^m\times a^n=a^{m+n}$]

$\therefore (-4x{y}^3)\times \left(6x^{2}y\right)=-24x^3{y}^4$----(1)

Now, lets verify for $x=2$ and $y=1$

LHS $=(-4x{y}^3)\times \left(6x^{2}y\right)$

$=(-4\times 2\times 1^3)(6\times 2^2\times 1)$

$=(-8)\times 24=-192$

RHS $=-24x^3{y}^4$

$=-24\times 2^3\times 1^4$

$=-24\times 8\times 1$

$=-192$

Therefore, LHS $=$ RHS

Hence, $(-4x{y}^3)\times \left(6x^{2}y\right)=-24x^3{y}^4$ verified.