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Chapter 0: Problem 109

Factor completely. $$x^{4}-5 x^{2} y^{2}+4 y^{4}$$

Short Answer

Expert verified
The solution to the problem \(x^{4}-5 x^{2} y^{2}+4 y^{4}\) when completely factorized is \((x^2 - 4y)(x^2 - y)\)

Step by step solution

01

Substitution

Substitute \(x^2\) with \(m\), we get: \(m^{2} - 5m*y + 4y^{2}\)
02

Factorize

Now we have a regular quadratic equation. Let's factorize the quadratic equation: \((m - 4y)(m - y)\)
03

Replace back

Substitute \(m\) with \(x^2\). We have the factors as: \((x^2 - 4y)(x^2 - y)\)

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Most popular questions from this chapter

What's wrong with this argument? Suppose \(x\) and \(y\) represent two real numbers, where \(x>y .\) $$\begin{aligned} &2>1\\\ &2(y-x)>1(y-x)\\\ &2 y-2 x>y-x\\\ &\begin{aligned} y-2 x &>-x \\ y &>x \end{aligned} \end{aligned}$$ The final inequality, \(y>x,\) is impossible because we were initially given \(x>y\)

Perform the indicated operations. $$(x-y)^{-1}+(x-y)^{-2}$$

Describe the solution set of \(|x|>-4\)

Explain how to solve \(x^{2}+6 x+8=0\) by completing the square.

Explain how to solve \(x^{2}+6 x+8=0\) using factoring and the zero-product principle.
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