Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-5x^{2}+13x+6$$
Factor the expression by grouping. First, the expression needs to be rewritten as $-5x^{2}+ax+bx+6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=13$$ $$ab=-5\times 6=-30$$
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 $-30$.
$$-1,30$$ $$-2,15$$ $$-3,10$$ $$-5,6$$
Calculate the sum for each pair.
$$-1+30=29$$ $$-2+15=13$$ $$-3+10=7$$ $$-5+6=1$$
The solution is the pair that gives sum $13$.
$$a=15$$ $$b=-2$$
Rewrite $-5x^{2}+13x+6$ as $\left(-5x^{2}+15x\right)+\left(-2x+6\right)$.
$$\left(-5x^{2}+15x\right)+\left(-2x+6\right)$$
Factor out $5x$ in the first and $2$ in the second group.
$$5x\left(-x+3\right)+2\left(-x+3\right)$$
Factor out common term $-x+3$ by using distributive property.
$$\left(-x+3\right)\left(5x+2\right)$$
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 $ax^{2}+bx+c=0$.
$$-5x^{2}+13x+6=0$$
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 $x=\frac{-13±17}{-10}$ when $±$ is plus. Add $-13$ to $17$.
$$x=\frac{4}{-10}$$
Reduce the fraction $\frac{4}{-10}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{2}{5}$$
Now solve the equation $x=\frac{-13±17}{-10}$ when $±$ is minus. Subtract $17$ from $-13$.
$$x=-\frac{30}{-10}$$
Divide $-30$ by $-10$.
$$x=3$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-\frac{2}{5}$ for $x_{1}$ and $3$ for $x_{2}$.
