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Intermediate Algebra

OPENSTAX

$Math Studyset Vaia Explanations$ Math

OER 2017 Edition · 1346 pages

ISBN: 9780998625720

Chapter 6: Q. 405 (page 636)

In the following exercises, factor completely
$24 x^{3} + 44 x^{2}$ .

Short Answer

Expert verified

The factors of the given expression are $4 x^{2} (6 x + 11)$ .

Step by step solution

01

Step 1. Given and explanation

We have $24 x^{3} + 44 x^{2}$ .

We can factorise it by taking common.

This will help in making the equation more small.

02

Step 2. Factorizing the expression.

We have $24 x^{3} + 44 x^{2}$ .

Factorising by taking common,

Therefore,

$24 x^{3} + 44 x^{2} = 4 x^{2} (6 x + 11)$

$24 x^{3} + 44 x^{2} = 4 x^{2} (6 x + 11)$

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Most popular questions from this chapter

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.
$8 y^{3} - 125 z^{3}$

In the following exercises, factor completely using the difference of squares pattern, if possible.
$x^{2} - 16 x + 64 - y^{2}$

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

$y^{2} - 5 y + 6$

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

u^{2} + 101 u + 100

Factor completely, $6 y^{3} + 18 y^{2} - 60 y$

See all solutions

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