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Proving Polynomial Identities

Proving Polynomial Identities 1.11 - Solution

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a
We have a positive sign inside the parentheses so we'll use the positive squared binomial. (a+b)2=a2+2ab+b2 (a+b)^2=a^2+2ab+b^2 In our expression (1+2a)2,(1+2a)^2, the variables are a=1a=1 and b=2a.b=2a.
(1+2a)2\left(1+2a\right)^2
12+212a+(2a)21^2+2\cdot 1\cdot 2a+(2a)^2
12+212a+22a21^2+2\cdot 1\cdot 2a+2^2a^2
Simplify power and product
1+4a+4a21+4a+4a^2
b
Again, we use the positive squared binomial, since there is a positive sign inside the parentheses.
(9+x2)2\left(9+x^2\right)^2
92+29x2+(x2)29^2+2\cdot 9\cdot x^2+\left(x^2\right)^2
92+29x2+x49^2+2\cdot 9\cdot x^2+x^4
Simplify power and product
81+18x2+x481+18x^2+x^4
c
Since we have a negative sign inside the parentheses, we use the second squared binomial: (ab)2=a22ab+b2. (a-b)^2=a^2-2ab+b^2. Now a=4ya=4y and b=x2.b=\frac{x}{2}. Remember to raise the entire term 4y4y and the entire term x2\frac{x}{2} by 22.
(4yx2)2\left(4y-\dfrac{x}{2}\right)^2
(4y)224yx2+(x2)2(4y)^2-2\cdot 4y\cdot \dfrac{x}{2}+\left(\dfrac{x}{2}\right)^2
42y224yx2+(x2)24^2y^2-2\cdot 4y\cdot \dfrac{x}{2}+\left(\dfrac{x}{2}\right)^2
42y224yx2+x2224^2y^2-2\cdot 4y\cdot \dfrac{x}{2}+\dfrac{x^2}{2^2}
Simplify power and product
16y28yx2+x2416y^2-8y\cdot \dfrac{x}{2}+\dfrac{x^2}{4}
16y28xy2+x2416y^2-\dfrac{8xy}{2}+\dfrac{x^2}{4}
16y24xy+x2416y^2-4xy+\dfrac{x^2}{4}