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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 6.47TI. (page 594)

Factor: $64 y^{2} - 80 y + 25$ .

Short Answer

Expert verified

Factors of $64 y^{2} - 80 y + 25$ is ${(8 y - 5)}^{2}$ , that is $(8 y - 5) (8 y - 5)$ .

Step by step solution

01

Step 1. Given information.

Given an expression $64 y^{2} - 80 y + 25$ .

02

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

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

Last term is 25 which is $5^{2}$ .

So, both first and last term are perfect squares.

Consider $a = 8 y$ and $b = 5$ .

03

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

Middle term is $- 80 y$ which can be expressed as $- 2 脳 8 y 脳 5$ .

It follows that middle term is of the form $- 2 a b$ .

04

Step 4. Write the square of the binomial.

So, $64 y^{2} - 80 y + 25$ 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, $64 y^{2} - 80 y + 25 = {(8 y - 5)}^{2}$ .

05

Step 5. Check by multiplying.

Verification can be obtained as follows.

$\begin{array}{rcl} {(8 y - 5)}^{2} & = & {(8 y)}^{2} - 2 脳 8 y 脳 5 + 5^{2} \\ = & 64 y^{2} - 80 y + 25 \end{array}$

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