We have a quadratic of the form $ax_{2}+bx+c.$ Since there are no common factors, we will rewrite the linear term $4x$ as the sum of two linear terms. The of these two terms will be factors of $ac.$
$4x_{2}−4x−35⇔4x_{2}+(4)x+(35) $
We have that $a=4,$ $b=4,$ and $c=35.$ There are now three steps we need to follow in order to rewrite the above expression.
 Find $ac.$ Since we have that $a=4$ and $c=35,$ the value of $ac$ is $4⋅(35)=140.$
 Find factors of $ac$. Since $ac=140,$ which is negative, we need factors of $ac$ to have opposite signs in order for the product to be negative. Since $b=4,$ which is also negative, the absolute value of the negative factor will need to be greater than the absolute value of the positive factor, so that their sum is negative.
$1_{st}factor 124510 2_{nd}factor 14070352814 Sum 1+(140)2+(70)4+(35)5+(28)10+(14) Result 1396831234 $

Rewrite $bd$ as two terms. Now that we know which factors are the ones to be used, we can rewrite $bd$ as two terms. $4x_{2}+(4)x−35⇔4x_{2}−14x+10x−35 $
Finally, we will factor the last expression obtained.
$4x_{2}−14x+10x−35$
$2x(2x−7)+10x−35$
$2x(2x−7)+5(2x−7)$
$(2x−7)(2x+5)$