Q 6.47TI. - Intermediate Algebra

Question

Factor: $64y^2-80y+25$.

Step-by-Step Solution

Answer

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

Step 1. Given information.

Given an expression $64y^2-80y+25$.

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

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

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

So, both first and last term are perfect squares.

Consider $a=8y$ and $b=5$.

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

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

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

Step 4. Write the square of the binomial.

So, $64y^2-80y+25$ is a perfect square trinomial of the form $a^2-2ab+b^2$.

Note that ${\left(a-b\right)}^2=a^2-2ab+b^2$.

Therefore, $64y^2-80y+25={\left(8y-5\right)}^2$.

Step 5. Check by multiplying.

Verification can be obtained as follows.

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