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Chapter 0: Problem 82

Factor completely, or state that the polynomial is prime. $$y^{5}-16 y$$

Short Answer

Expert verified
The factored form of \(y^{5}-16 y\) is \(y(y - 2)(y + 2)(y^{2} + 4)\).

Step by step solution

01

Identify the common factor

Firstly, identify the common factor in both terms. Here, the common factor is \(y\).
02

Factor out the common factor

Next, factor out the common factor (\(y\)). This leaves you with \(y(y^{4} - 16)\).
03

Factor the quadratic portion

Try to factor the remaining part, \(y^{4} - 16\), which is the difference of two squares. This can further be factored into \((y^{2} - 4)(y^{2} + 4)\).
04

Factoring completely

Finally, the term \(y^{2} - 4\) is also a difference of squares and can be further factored into \((y - 2)(y + 2)\). The term \(y^{2} + 4\) cannot be further factored.

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Most popular questions from this chapter

What does it mean to factor completely?

$$\text { Factor completely.}$$ $$2 x^{2}-7 x y^{2}+3 y^{4}$$

$$\text { factor completely.}$$ $$(x-5)^{-\frac{1}{2}}(x+5)^{-\frac{1}{2}}-(x+5)^{\frac{1}{2}}(x-5)^{-\frac{3}{2}}$$

$$\text { factor completely.}$$ $$-x^{2}-4 x+5$$

Multiply: \(\quad\left(2 x^{3} y^{2}\right)\left(5 x^{4} y^{7}\right)\)
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