Question

Factor $25v^2+20v+4$completely using the perfect square trinomials pattern.

Short answer

Factors of$25v^2+20v+4$are${\left(5v+2\right)}^2$.

Step-by-step solution

Step 1. Given information.

Given an expression $25v^2+20v+4$.

Step 2. Check if the first and last term of the expression a perfect square.

First term and last term are $25v^2$and 4 respectively.

Consider $a=5v$and $b=2$.

It can be seen that $25v^2={\left(5v\right)}^2$and $4=2^2$.

Therefore, first and last terms are perfect squares.

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

Middle term of the expression is $20v$.

Also, $20v=2\left(5v\right)\left(2\right)$.

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

Step 4. Write the square of the binomial.

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

Obtain the result as follows.

$\begin{array}{rcl}25v^2+20v+4& =& {\left(5v\right)}^2+2\left(5v\right)2+2^2\\ & =& {\left(5v+2\right)}^2\end{array}$

Step 5. Check the result.

Verify the result as follows.

$\begin{array}{rcl}{\left(5v+2\right)}^2& =& {\left(5v\right)}^2+2\left(5v\right)2+2^2\\ & =& 25v^2+20v+4\end{array}$