Q 92 In the following exercises, fact... [FREE SOLUTION]

91影视

Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 6: Q 92 (page 589)

In the following exercises, factor completely using trial and error.
$11 n^{3} - 55 n^{2} + 44 n$

Short Answer

Expert verified

The factorization of the expression is $11 n^{3} - 55 n^{2} + 44 n = 11 n (n - 1) (n - 4)$

Step by step solution

Step 1. Given Information

Consider the expression $11 n^{3} - 55 n^{2} + 44 n$

The objective is to factor the expression completely by using the trial and error method.

Step 2. Factor out the GCF

The GCF of the three terms in the expression is $11 n$ . So factor the GCF from the expression we get

$\begin{array}{rcl} 11 n^{3} - 55 n^{2} + 44 n & = & 11 n 路 n^{2} - 11 n 路 5 n + 11 n 路 4 \\ = & 11 n (n^{2} - 5 n + 4) \end{array}$

Step 3. Factor using trial and error

The factors of the first term and the last term of the trinomial $n^{2} - 5 n + 4$

$n^{2} = n 路 n 4 = 2 路 2 o r 4 路 1$

Consider all the combinations of factors

$n^{2} - 5 n + 4$
Possible Factors	Product
$(n - 2) (n - 2)$	$n^{2} - 2 n - 2 n + 4 = n^{2} - 4 n + 4$
$(n + 2) (n + 2)$	$n^{2} + 2 n + 2 n + 4 = n^{2} + 4 n + 4$
$(n - 4) (n - 1)$	$n^{2} - n - 4 n + 4 = n^{2} - 5 n + 4$
$(n + 4) (n + 1)$	$n^{2} + n + 4 n + 4 = n^{2} + 5 n + 4$

So the other factors are $(n - 4) (n - 1)$ .

Thus, $11 n^{3} - 55 n^{2} + 44 n = 11 n (n - 1) (n - 4)$

Step 4. Check the factors

Multiply the factors to check the solution

$\begin{array}{rcl} 11 n (n - 1) (n - 4) & = & 11 n (n^{2} - 4 n - n + 4) \\ = & 11 n (n^{2} - 5 n + 4) \\ = & 11 n^{3} - 55 n^{2} + 44 n \end{array}$

And we get the given expression. So the expression is correctly factored.

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91影视

In the following exercises, factor completely using trial and error.
$11 n^{3} - 55 n^{2} + 44 n$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Factor out the GCF

Step 3. Factor using trial and error

Step 4. Check the factors

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91影视

In the following exercises, factor completely using trial and error.11n3-55n2+44n

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Factor out the GCF

Step 3. Factor using trial and error

Step 4. Check the factors

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Mechanics Maths

Pure Maths

Discrete Mathematics

Probability and Statistics

Calculus

Study anywhere. Anytime. Across all devices.

In the following exercises, factor completely using trial and error.
$11 n^{3} - 55 n^{2} + 44 n$