Given:
[tex](y-4 x)\left(y^{2}+4 y+16\right)[/tex]
Resulting form = [tex]y^{3}+4 y^{2}+a y-4 x y^{2}-a x y-64 x[/tex]
To find:
The value of a in the polynomial.
Solution:
[tex](y-4 x)\left(y^{2}+4 y+16\right)[/tex]
Using distributive property:
[tex]a(b+c)=a b+a c[/tex]
[tex](y-4 x)(y^{2}+4 y+16)=y(y^{2}+4 y+16)-4 x(y^{2}+4 y+16)[/tex]
Now multiply each of the first term with each of the second term.
[tex]=y y^{2}+y \cdot 4 y+y \cdot 16+(-4 x) y^{2}+(-4 x) \cdot 4 y+(-4 x) \cdot 16[/tex]
Applying plus minus rule: [tex]+(-a)=-a[/tex]
[tex]=y^{2} \cdot y+4 y\cdot y+16 y-4 y^{2} x-4 \cdot 4 y x-4 \cdot 16 x[/tex]
Apply the exponent rule: [tex]a^{b} \cdot a^{c}=a^{b+c}[/tex]
[tex]=y^{3}+4 y^2+16 y-4 y^{2} x-16 y x-64 x[/tex]
Let us compare this with the given resulting form:
[tex]=y^{3}+4 y^{2}+a y-4 x y^{2}-a x y-64 x[/tex]
On comparing above two expression, we get a = 16.
The value of a in the polynomial is 16.
