Question

Multiplying polynomials and simplifying expressions

Answer 1

Given:

$(y-4 x)\left(y^{2}+4 y+16\right)$

Resulting form = $y^{3}+4 y^{2}+a y-4 x y^{2}-a x y-64 x$

To find:

The value of a in the polynomial.

Solution:

$(y-4 x)\left(y^{2}+4 y+16\right)$

Using distributive property:

$a(b+c)=a b+a c$

$(y-4 x)(y^{2}+4 y+16)=y(y^{2}+4 y+16)-4 x(y^{2}+4 y+16)$

Now multiply each of the first term with each of the second term.

                     $=y y^{2}+y \cdot 4 y+y \cdot 16+(-4 x) y^{2}+(-4 x) \cdot 4 y+(-4 x) \cdot 16$

Applying plus minus rule: $+(-a)=-a$

                    $=y^{2} \cdot y+4 y\cdot y+16 y-4 y^{2} x-4 \cdot 4 y x-4 \cdot 16 x$

Apply the exponent rule: $a^{b} \cdot a^{c}=a^{b+c}$

                    $=y^{3}+4 y^2+16 y-4 y^{2} x-16 y x-64 x$

Let us compare this with the given resulting form:

                   $=y^{3}+4 y^{2}+a y-4 x y^{2}-a x y-64 x$

On comparing above two expression, we get a = 16.

The value of a in the polynomial is 16.