Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-8x^{2}-10x+3$$
Factor the expression by grouping. First, the expression needs to be rewritten as $-8x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-10$$ $$ab=-8\times 3=-24$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. 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 $-10$.
$$a=2$$ $$b=-12$$
Rewrite $-8x^{2}-10x+3$ as $\left(-8x^{2}+2x\right)+\left(-12x+3\right)$.
$$\left(-8x^{2}+2x\right)+\left(-12x+3\right)$$
Factor out $2x$ in the first and $3$ in the second group.
$$2x\left(-4x+1\right)+3\left(-4x+1\right)$$
Factor out common term $-4x+1$ by using distributive property.
$$\left(-4x+1\right)\left(2x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$-8x^{2}-10x+3=0$$
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 $x=\frac{10±14}{-16}$ when $±$ is plus. Add $10$ to $14$.
$$x=\frac{24}{-16}$$
Reduce the fraction $\frac{24}{-16}$ to lowest terms by extracting and canceling out $8$.
$$x=-\frac{3}{2}$$
Now solve the equation $x=\frac{10±14}{-16}$ when $±$ is minus. Subtract $14$ from $10$.
$$x=-\frac{4}{-16}$$
Reduce the fraction $\frac{-4}{-16}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{1}{4}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-\frac{3}{2}$ for $x_{1}$ and $\frac{1}{4}$ for $x_{2}$.
Multiply $\frac{-2x-3}{-2}$ times $\frac{-4x+1}{-4}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
