Let's start by factoring out the . Then, we will factor the resulting .
Factor Out the Greatest Common Factor
The greatest common factor is a common factor of all the terms in the expression. It is the common factor with the greatest and the greatest . In this case, the greatest common factor is
$x_{2}.$
$2x_{4}+7x_{3}+5x_{2}$
$x_{2}⋅2x_{2}+x_{2}⋅7x+x_{2}⋅5$
$x_{2}(2x_{2}+7x+5)$
The result of factoring out the greatest common factor from the given expression is a with a of
$2.$
$x_{2}(2x_{2}+7x+5) $
Let's temporarily only focus on the expression in parentheses, and bring back the greatest common factor after factoring.
Factor the Quadratic Trinomial
Here we have a quadratic of the form $ax_{2}+bx+c,$ where $∣a∣ =1$ and there are no common factors. To factor this expression, we will rewrite the middle term, $bx,$ as two terms. The coefficients of these two terms will be factors of $ac$ whose sum must be $b.$
$2x_{2}+7x+5 $
We have that $a=2,$ $b=7,$ and $c=5.$ There are now three steps we need to follow in order to rewrite the above expression.
 Find $ac.$ Since we have that $a=2$ and $c=5,$ the value of $ac$ is $2×5=10.$
 Find factors of $ac$. Since $ac=10,$ which is positive, we need factors of $ac$ to have the same sign — both positive or both negative — in order for the product to be positive. Since $b=7,$ which is also positive, those factors will need to be positive so that their sum is positive.
$1_{st}factor 22 2_{nd}factor 55 Sum 2+(5)2+5 Result 77 $

Rewrite $bx$ as two terms. Now that we know which factors are the ones to be used, we can rewrite $bx$ as two terms. $2x_{2}+7x+5⇔2x_{2}+2x+5x+5 $
Finally, we will factor the last expression obtained.
$2x_{2}+2x+5x+5$
$2x(x+1)+5x+5$
$2x(x+1)+5(x+1)$
$(x+1)(2x+5)$
We have factored the quadratic trinomial.
$2x_{2}+7x+5⇔(x+1)(2x+5) $
Finally, we need to include the greatest common factor that we factored out at the beginning.
$2x_{4}+7x_{3}+5x_{2}⇔x_{2}(x+1)(2x+5) $