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Intermediate Algebra

OPENSTAX

$Math Studyset Vaia Explanations$ Math

OER 2017 Edition · 1346 pages

ISBN: 9780998625720

Chapter 6: Q. 235 (page 613)

In the following exercises, factor completely.
$a^{5} + 9 a^{3}$

Short Answer

Expert verified

The required factors are $a^{3} (a^{2} + 9)$

Step by step solution

01

Step 1. Given Information

The given expression is $a^{5} + 9 a^{3}$

02

Step 2. Calculation

We will find the factors by taking the common values outside the paranthesis.

$a^{5} + 9 a^{3} = a^{3} (a^{2} + 9)$

03

Step 3. Check

We will verify the factors by multiplying the factors.

$a^{3} (a^{2} + 9) = a^{5} + 9 a^{3}$

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Most popular questions from this chapter

Factor completely using trial and error:

$56 q^{3} + 320 q^{2} - 96 q$

In the following exercises, factor each trinomial of the form $x^{2} + b x + c$ .

$m^{2} - 13 m + 30$

Factor completely using trial and error:

$15 n^{3} - 85 n^{2} + 100 n$

Factor completely using trial and error:

$30 x^{2} - 53 x y - 21 y^{2}$

Factor Trinomials of the Form $a x^{2} + b x + c$ Using Trial and Error.

$4 w^{2} - 5 w + 1$

See all solutions

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