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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 6: Q 6.50 (page 595)

Factor: $64 m^{2} + 112 m n + 49 n^{2}$ .

Short Answer

Expert verified

Factors of $64 m^{2} + 112 m n + 49 n^{2}$ is ${(8 m + 7 n)}^{2}$ .

Step by step solution

01

Step 1. Given information.

Given an expression $64 m^{2} + 112 m n + 49 n^{2}$ .

02

Step 2. Find if first and last term are perfect squares or not.

First term is $64 m^{2}$ which can be expressed as ${(8 m)}^{2}$ .

Last term is $49 n^{2}$ which can be expressed as ${(7 n)}^{2}$ .

Consider $a = 8 m$ and $b = 7 n$ .

So, first and last term are perfect squares.

03

Step 3. Check if the middle term is of the form ±2ab.

Here, middle term is $112 m n$ which can be expressed as $2 脳 8 m 脳 7 n$ .

It follows that middle term is of the form .

04

Step 4. Write the square of the binomial.

So, $64 m^{2} + 112 m n + 49 n^{2}$ is a perfect square trinomial of the form $a^{2} + 2 a b + b^{2}$ .

Note that ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ .

Therefore, $64 m^{2} + 112 m n + 49 n^{2} = {(8 m + 7 n)}^{2}$ .

05

Step 5. Check by multiplying.

Verification can be obtained as follows.

$\begin{array}{rcl} {(8 m + 7 n)}^{2} & = & {(8 m)}^{2} + 2 脳 8 m 脳 7 n + {(7 n)}^{2} \\ = & 64 m^{2} + 112 m n + 49 n^{2} \end{array}$

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Most popular questions from this chapter

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