Consider $2x^{2}-5x-3$. Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-5$$ $$ab=2\left(-3\right)=-6$$
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 $-6$.
$$1,-6$$ $$2,-3$$
Calculate the sum for each pair.
$$1-6=-5$$ $$2-3=-1$$
The solution is the pair that gives sum $-5$.
$$a=-6$$ $$b=1$$
Rewrite $2x^{2}-5x-3$ as $\left(2x^{2}-6x\right)+\left(x-3\right)$.
$$\left(2x^{2}-6x\right)+\left(x-3\right)$$
Factor out $2x$ in $2x^{2}-6x$.
$$2x\left(x-3\right)+x-3$$
Factor out common term $x-3$ by using distributive property.
$$\left(x-3\right)\left(2x+1\right)$$
Rewrite the complete factored expression.
$$3\left(x-3\right)\left(2x+1\right)$$
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 $ax^{2}+bx+c=0$.
$$6x^{2}-15x-9=0$$
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 $x=\frac{15±21}{12}$ when $±$ is plus. Add $15$ to $21$.
$$x=\frac{36}{12}$$
Divide $36$ by $12$.
$$x=3$$
Now solve the equation $x=\frac{15±21}{12}$ when $±$ is minus. Subtract $21$ from $15$.
$$x=-\frac{6}{12}$$
Reduce the fraction $\frac{-6}{12}$ to lowest terms by extracting and canceling out $6$.
$$x=-\frac{1}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $3$ for $x_{1}$ and $-\frac{1}{2}$ for $x_{2}$.
