Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-2y^{2}-y+6$$
Factor the expression by grouping. First, the expression needs to be rewritten as $-2y^{2}+ay+by+6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-1$$ $$ab=-2\times 6=-12$$
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 $-12$.
$$1,-12$$ $$2,-6$$ $$3,-4$$
Calculate the sum for each pair.
$$1-12=-11$$ $$2-6=-4$$ $$3-4=-1$$
The solution is the pair that gives sum $-1$.
$$a=3$$ $$b=-4$$
Rewrite $-2y^{2}-y+6$ as $\left(-2y^{2}+3y\right)+\left(-4y+6\right)$.
$$\left(-2y^{2}+3y\right)+\left(-4y+6\right)$$
Factor out $-y$ in the first and $-2$ in the second group.
$$-y\left(2y-3\right)-2\left(2y-3\right)$$
Factor out common term $2y-3$ by using distributive property.
$$\left(2y-3\right)\left(-y-2\right)$$
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 $ax^{2}+bx+c=0$.
$$-2y^{2}-y+6=0$$
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 $y=\frac{1±7}{-4}$ when $±$ is plus. Add $1$ to $7$.
$$y=\frac{8}{-4}$$
Divide $8$ by $-4$.
$$y=-2$$
Now solve the equation $y=\frac{1±7}{-4}$ when $±$ is minus. Subtract $7$ from $1$.
$$y=-\frac{6}{-4}$$
Reduce the fraction $\frac{-6}{-4}$ to lowest terms by extracting and canceling out $2$.
$$y=\frac{3}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-2$ for $x_{1}$ and $\frac{3}{2}$ for $x_{2}$.
