Split the second term in \(2{x}^{2}+5x-12\) into two terms.
Multiply the coefficient of the first term by the constant term.
\[2\times -12=-24\]
Ask: Which two numbers add up to \(5\) and multiply to \(-24\)?
Split \(5x\) as the sum of \(8x\) and \(-3x\).
\[2{x}^{2}+8x-3x-12\]
\[\begin{aligned}&n\\&\frac{2{x}^{2}+8x-3x-12}{4{x}^{2}-9}\end{aligned}\]
Factor out common terms in the first two terms, then in the last two terms.
\[\begin{aligned}&n\\&\frac{2x(x+4)-3(x+4)}{4{x}^{2}-9}\end{aligned}\]
Factor out the common term \(x+4\).
\[\begin{aligned}&n\\&\frac{(x+4)(2x-3)}{4{x}^{2}-9}\end{aligned}\]
Rewrite \(4{x}^{2}-9\) in the form \({a}^{2}-{b}^{2}\), where \(a=2x\) and \(b=3\).
\[\begin{aligned}&n\\&\frac{(x+4)(2x-3)}{{(2x)}^{2}-{3}^{2}}\end{aligned}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\begin{aligned}&n\\&\frac{(x+4)(2x-3)}{(2x+3)(2x-3)}\end{aligned}\]
n;((x+4)*(2*x-3))/((2*x+3)*(2*x-3))
