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Chapter 0: Problem 81

Factor completely, or state that the polynomial is prime. $$y^{5}-81 y$$

Short Answer

Expert verified
The fully factored form of \(y^{5}-81 y\) is \( y(y^{2} + 9)( y+3 )(y-3) \).

Step by step solution

01

Identify Common Factors

In the given polynomial \(y^{5}-81y\), the common factor is \(y\).
02

Remove the Common Factor

Remove the common factor by dividing both terms in the binomial by y to get \(y(y^{4}- 81)\). Now, the polynomial has been simplified.
03

Recognize the Difference of Squares

In the term within the brackets, the expression \(y^{4}- 81\) can be written as \( (y^{2})^{2} - (9)^{2} \), which is a difference of squares.
04

Use the Identity of Difference of Squares

Apply the identity \(a^2 - b^2 = (a+b)(a-b)\). By applying this identity where \( a = y^{2}\) and \( b = 9\), we can rewrite the above expression as \( (y^{2}+9)(y^{2}-9) \).
05

Use the Identity Again

The expression \( y^{2} - 9 \) is again a difference of squares with \( a = y \) and \( b = 3 \) so it can be further factored as \( y+3 \) and \( y-3 \).
06

Write the Final Factored Form

Now, substitute these factors back into the expression. The fully factored form of the given binomial is \( y(y^{2} + 9)( y+3 )(y-3) \).

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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 \(x>y\)

Will help you prepare for the material covered in the first section of the next chapter. If \(y=4-x^{2},\) find the value of \(y\) that corresponds to values of \(x\) for each integer starting with -3 and ending with 3

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