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Chapter 2: Problem 19

Factor \(x^{8}-y^{8}\) as nicely as possible.

Short Answer

Expert verified
The short answer to factoring \(x^8 - y^8\) is: \((x^8 - y^8) = (x^4 + y^4)(x^2 + y^2)(x^2 - y^2)\)

Step by step solution

01

Rewrite the expression as a difference of squares

We can rewrite \(x^8 - y^8\) as a difference of squares, that is \((x^4)^2 - (y^4)^2\). Now apply the difference of squares formula, which states that \(a^2 - b^2 = (a + b)(a - b)\). In this case, \(a = x^4\) and \(b = y^4\). So we have: \[(x^4)^2 - (y^4)^2 = (x^4 + y^4)(x^4 - y^4)\]
02

Factor the difference of squares further, if possible

Now, let's check if we can factor the difference of squares further. We can rewrite \((x^4 - y^4)\) as another difference of squares, that is \((x^2)^2 - (y^2)^2\). Again applying the difference of squares formula, with \(a = x^2\) and \(b = y^2\), we get: \[(x^2)^2 - (y^2)^2 = (x^2 + y^2)(x^2 - y^2)\]
03

Combine the factors obtained in steps 1 and 2

We have obtained the following factors for our original expression from steps 1 and 2: \((x^4 + y^4)(x^4 - y^4)\) and \((x^2 + y^2)(x^2 - y^2)\) Now combine these factors as the factorization of the original expression: \[(x^8 - y^8) = (x^4 + y^4)(x^2 + y^2)(x^2 - y^2)\]
04

Check to ensure we can't factor any further

Notice that \(x^2 + y^2\) and \(x^4 + y^4\) cannot be factored further using the difference of squares or difference of cubes formulas, because they are sums, not differences. So we are left with the most simplified factorization of the expression \(x^8 - y^8\): \[(x^8 - y^8) = (x^4 + y^4)(x^2 + y^2)(x^2 - y^2)\]

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Most popular questions from this chapter

A bicycle company finds that its average cost per bicycle for producing \(n\) thousand bicycles is \(a(n)\) dollars, where $$ a(n)=700 \frac{4 n^{2}+3 n+50}{16 n^{2}+3 n+35} $$ What will be the approximate cost per bicycle when the company is producing many bicycles?

Show that if \(p\) and \(q\) are nonzero polynomials with \(\operatorname{deg} p<\operatorname{deg} q,\) then \(\operatorname{deg}(p+q)=\operatorname{deg} q\).

Write each expression as the sum of a polynomial and a rational function whose numerator has smaller degree than its denominator. $$ \frac{x^{6}-4 x^{2}+5}{x^{2}-3 x+1} $$

Write the indicated expression as a ratio of polynomials, assuming that $$ r(x)=\frac{3 x+4}{x^{2}+1}, \quad s(x)=\frac{x^{2}+2}{2 x-1}, \quad t(x)=\frac{5}{4 x^{3}+3} $$. $$ (s \circ r)(x) $$

Suppose \(t\) is a zero of the polynomial \(p\) defined by $$ p(x)=3 x^{5}+7 x^{4}+2 x+6 $$ Show that \(\frac{1}{t}\) is a zero of the polynomial \(q\) defined by $$ q(x)=3+7 x+2 x^{4}+6 x^{5} $$.
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