# Solve ((3x+5+x^2)/(2x-1))^4(-2x-13+2x^2)/(-4x+1+4x^2)*5

## 5\left(\frac{x^2+3x+5}{2x-1}\right)^4\cdot\frac{2x^2-2x-13}{4x^2-4x+1}

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$5\left(\frac{2\left(\left(\frac{3}{2}+x\right)^2+\frac{11}{4}\right)^4\left(\left(x-\frac{1}{2}\right)^2-\frac{27}{4}\right)}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$

## Step by step solution

Problem

$5\left(\frac{x^2+3x+5}{2x-1}\right)^4\cdot\frac{2x^2-2x-13}{4x^2-4x+1}$
1

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$5\frac{-13-2x+2x^2}{1-4x+4x^2}\cdot\frac{\left(5+3x+x^2\right)^4}{\left(2x-1\right)^4}$
2

Multiplying fractions

$5\left(\frac{\left(-13-2x+2x^2\right)\left(5+3x+x^2\right)^4}{\left(1-4x+4x^2\right)\left(2x-1\right)^4}\right)$
3

The trinomial $\left(1-4x+4x^2\right)\left(2x-1\right)^4$ is perfect square, because it's discriminant is equal to zero

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

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{4x^2}\:and\:b=\sqrt{1}$
5

Factoring the perfect square trinomial

$5\left(\frac{\left(-13-2x+2x^2\right)\left(5+3x+x^2\right)^4}{\left(2x-1\right)^{2}\left(2x-1\right)^4}\right)$
6

When multiplying exponents with same base we can add the exponents

$5\left(\frac{\left(-13-2x+2x^2\right)\left(5+3x+x^2\right)^4}{\left(2x-1\right)^{6}}\right)$
7

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

$5\left(\frac{\left(-13-2x+2x^2\right)\left(5+3x+x^2\right)^4}{\left(\left(\sqrt{2}\sqrt{x}-1\right)\left(1+\sqrt{2}\sqrt{x}\right)\right)^{6}}\right)$
8

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

$5\left(\frac{\left(-13-2x+2x^2\right)\left(5+3x+x^2\right)^4}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$
9

Use the complete the square method to factor the trinomial of the form $ax^2+bx+c$. Take common factor $a$ ($2$) to all terms

$5\left(\frac{2\left(5+3x+x^2\right)^4\left(-\frac{13}{2}-x+x^2\right)}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$
10

Add and subtract $\displaystyle\left(\frac{b}{2a}\right)^2$

$5\left(\frac{2\left(5+3x+x^2\right)^4\left(-\frac{1}{4}+\frac{1}{4}-\frac{13}{2}-x+x^2\right)}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$
11

Factor the perfect square trinomial $x^2+-1x+\frac{1}{4}$

$5\left(\frac{2\left(5+3x+x^2\right)^4\left(-\frac{1}{4}-\frac{13}{2}+\left(x-\frac{1}{2}\right)^2\right)}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$
12

Subtract the values $-\frac{13}{2}$ and $-\frac{1}{4}$

$5\left(\frac{2\left(5+3x+x^2\right)^4\left(\left(x-\frac{1}{2}\right)^2-\frac{27}{4}\right)}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$
13

Add and subtract $\displaystyle\left(\frac{b}{2a}\right)^2$

$5\left(\frac{2\left(-\frac{9}{4}+\frac{9}{4}+5+3x+x^2\right)^4\left(\left(x-\frac{1}{2}\right)^2-\frac{27}{4}\right)}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$
14

Factor the perfect square trinomial $x^2+3x+\frac{9}{4}$

$5\left(\frac{2\left(-\frac{9}{4}+5+\left(\frac{3}{2}+x\right)^2\right)^4\left(\left(x-\frac{1}{2}\right)^2-\frac{27}{4}\right)}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$
15

Subtract the values $5$ and $-\frac{9}{4}$

$5\left(\frac{2\left(\left(\frac{3}{2}+x\right)^2+\frac{11}{4}\right)^4\left(\left(x-\frac{1}{2}\right)^2-\frac{27}{4}\right)}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$

$5\left(\frac{2\left(\left(\frac{3}{2}+x\right)^2+\frac{11}{4}\right)^4\left(\left(x-\frac{1}{2}\right)^2-\frac{27}{4}\right)}{\left(\sqrt{2}\sqrt{x}-1\right)^{6}\left(1+\sqrt{2}\sqrt{x}\right)^{6}}\right)$

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