Factor the expression by grouping. First, the expression needs to be rewritten as $4x^{2}+ax+bx-6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=5$$ $$ab=4\left(-6\right)=-24$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-24$.
$$-1,24$$ $$-2,12$$ $$-3,8$$ $$-4,6$$
Calculate the sum for each pair.
$$-1+24=23$$ $$-2+12=10$$ $$-3+8=5$$ $$-4+6=2$$
The solution is the pair that gives sum $5$.
$$a=-3$$ $$b=8$$
Rewrite $4x^{2}+5x-6$ as $\left(4x^{2}-3x\right)+\left(8x-6\right)$.
$$\left(4x^{2}-3x\right)+\left(8x-6\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(4x-3\right)+2\left(4x-3\right)$$
Factor out common term $4x-3$ by using distributive property.
$$\left(4x-3\right)\left(x+2\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$4x^{2}+5x-6=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
Now solve the equation $x=\frac{-5±11}{8}$ when $±$ is plus. Add $-5$ to $11$.
$$x=\frac{6}{8}$$
Reduce the fraction $\frac{6}{8}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{3}{4}$$
Now solve the equation $x=\frac{-5±11}{8}$ when $±$ is minus. Subtract $11$ from $-5$.
$$x=-\frac{16}{8}$$
Divide $-16$ by $8$.
$$x=-2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{3}{4}$ for $x_{1}$ and $-2$ for $x_{2}$.
