Solve 2y^2+5y-12

$$2 { y }^{ 2 } +5y-12$$

Factor

$$\left(2y-3\right)\left(y+4\right)$$

Solution Steps

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

$$a+b=5 ab=2\left(-12\right)=-24$$

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 -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 5.

$$a=-3 b=8$$

Rewrite 2y^{2}+5y-12 as \left(2y^{2}-3y\right)+\left(8y-12\right).

$$\left(2y^{2}-3y\right)+\left(8y-12\right)$$

Factor out y in the first and 4 in the second group.

$$y\left(2y-3\right)+4\left(2y-3\right)$$

Factor out common term 2y-3 by using distributive property.

$$\left(2y-3\right)\left(y+4\right)$$

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.

$$2y^{2}+5y-12=0$$

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.

$$y=\frac{-5±\sqrt{5^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\times 2\left(-12\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-5±\sqrt{25-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$y=\frac{-5±\sqrt{25+96}}{2\times 2}$$

Add 25 to 96.

$$y=\frac{-5±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$y=\frac{-5±11}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-5±11}{4}$$

Now solve the equation y=\frac{-5±11}{4} when ± is plus. Add -5 to 11.

$$y=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$y=\frac{3}{2}$$

Now solve the equation y=\frac{-5±11}{4} when ± is minus. Subtract 11 from -5.

$$y=-\frac{16}{4}$$

Divide -16 by 4.

$$y=-4$$

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 -4 for x_{2}.

$$2y^{2}+5y-12=2\left(y-\frac{3}{2}\right)\left(y-\left(-4\right)\right)$$

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

$$2y^{2}+5y-12=2\left(y-\frac{3}{2}\right)\left(y+4\right)$$

Subtract \frac{3}{2} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2y^{2}+5y-12=2\times \frac{2y-3}{2}\left(y+4\right)$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2y^{2}+5y-12=\left(2y-3\right)\left(y+4\right)$$

Evaluate

$$\left(2y-3\right)\left(y+4\right)$$