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# Proving Polynomial Identities

## Proving Polynomial Identities 1.11 - Solution

a
We have a positive sign inside the parentheses so we'll use the positive squared binomial. $(a+b)^2=a^2+2ab+b^2$ In our expression $(1+2a)^2,$ the variables are $a=1$ and $b=2a.$
$\left(1+2a\right)^2$
$1^2+2\cdot 1\cdot 2a+(2a)^2$
$1^2+2\cdot 1\cdot 2a+2^2a^2$
Simplify power and product
$1+4a+4a^2$
b
Again, we use the positive squared binomial, since there is a positive sign inside the parentheses.
$\left(9+x^2\right)^2$
$9^2+2\cdot 9\cdot x^2+\left(x^2\right)^2$
$9^2+2\cdot 9\cdot x^2+x^4$
Simplify power and product
$81+18x^2+x^4$
c
Since we have a negative sign inside the parentheses, we use the second squared binomial: $(a-b)^2=a^2-2ab+b^2.$ Now $a=4y$ and $b=\frac{x}{2}.$ Remember to raise the entire term $4y$ and the entire term $\frac{x}{2}$ by $2$.
$\left(4y-\dfrac{x}{2}\right)^2$
$(4y)^2-2\cdot 4y\cdot \dfrac{x}{2}+\left(\dfrac{x}{2}\right)^2$
$4^2y^2-2\cdot 4y\cdot \dfrac{x}{2}+\left(\dfrac{x}{2}\right)^2$
$4^2y^2-2\cdot 4y\cdot \dfrac{x}{2}+\dfrac{x^2}{2^2}$
Simplify power and product
$16y^2-8y\cdot \dfrac{x}{2}+\dfrac{x^2}{4}$
$16y^2-\dfrac{8xy}{2}+\dfrac{x^2}{4}$
$16y^2-4xy+\dfrac{x^2}{4}$