Solve ((1-1x^2)^2)/(2x+1+x^2)

\frac{\left(1-x^2\right)^2}{x^2+2x+1}

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Answer

$\left(1-x\right)^2$

Step by step solution

Problem

$\frac{\left(1-x^2\right)^2}{x^2+2x+1}$
1

The trinomial $\frac{\left(1-x^2\right)^2}{1+2x+x^2}$ is perfect square, because it's discriminant is equal to zero

$\Delta=b^2-4ac=2^2-4\left(1\right)\left(1\right) = 0$
2

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{x^2}\:and\:b=\sqrt{1}$
3

Factoring the perfect square trinomial

$\frac{\left(1-x^2\right)^2}{\left(1+x\right)^{2}}$
4

Rewrite the difference of squares $\left(1-x^2\right)$ as the product of two conjugated binomials

$\frac{\left(\left(1-x\right)\left(x+1\right)\right)^2}{\left(1+x\right)^{2}}$
5

The power of a product is equal to the product of it's factors raised to the same power

$\frac{\left(1-x\right)^2\left(x+1\right)^2}{\left(1+x\right)^{2}}$
6

Simplifying the fraction by $\left(1+x\right)^{2}$

$\left(1-x\right)^2$

Answer

$\left(1-x\right)^2$

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Problem Analysis

Main topic:

Factor by difference of squares

Time to solve it:

0.26 seconds

Views:

139