Solve y^2-y-12

$${ y }^{ 2 } -y-12$$

Factor

$$\left(y-4\right)\left(y+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-12. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-12\right)=-12$$

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.

$$a=-4 b=3$$

Rewrite y^{2}-y-12 as \left(y^{2}-4y\right)+\left(3y-12\right).

$$\left(y^{2}-4y\right)+\left(3y-12\right)$$

Factor out y in the first and 3 in the second group.

$$y\left(y-4\right)+3\left(y-4\right)$$

Factor out common term y-4 by using distributive property.

$$\left(y-4\right)\left(y+3\right)$$

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.

$$y^{2}-y-12=0$$

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.

$$y=\frac{-\left(-1\right)±\sqrt{1-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$y=\frac{-\left(-1\right)±\sqrt{1+48}}{2}$$

Add 1 to 48.

$$y=\frac{-\left(-1\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$y=\frac{-\left(-1\right)±7}{2}$$

The opposite of -1 is 1.

$$y=\frac{1±7}{2}$$

Now solve the equation y=\frac{1±7}{2} when ± is plus. Add 1 to 7.

$$y=\frac{8}{2}$$

Divide 8 by 2.

$$y=4$$

Now solve the equation y=\frac{1±7}{2} when ± is minus. Subtract 7 from 1.

$$y=-\frac{6}{2}$$

Divide -6 by 2.

$$y=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and -3 for x_{2}.

$$y^{2}-y-12=\left(y-4\right)\left(y-\left(-3\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$y^{2}-y-12=\left(y-4\right)\left(y+3\right)$$

Evaluate

$$\left(y-4\right)\left(y+3\right)$$