Solve 6y^2+23y-18

$$6 { y }^{ 2 } +23y-18$$

Factor

$$\left(3y-2\right)\left(2y+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6y^{2}+ay+by-18. To find a and b, set up a system to be solved.

$$a+b=23 ab=6\left(-18\right)=-108$$

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 -108.

$$-1,108 -2,54 -3,36 -4,27 -6,18 -9,12$$

Calculate the sum for each pair.

$$-1+108=107 -2+54=52 -3+36=33 -4+27=23 -6+18=12 -9+12=3$$

The solution is the pair that gives sum 23.

$$a=-4 b=27$$

Rewrite 6y^{2}+23y-18 as \left(6y^{2}-4y\right)+\left(27y-18\right).

$$\left(6y^{2}-4y\right)+\left(27y-18\right)$$

Factor out 2y in the first and 9 in the second group.

$$2y\left(3y-2\right)+9\left(3y-2\right)$$

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

$$\left(3y-2\right)\left(2y+9\right)$$

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.

$$6y^{2}+23y-18=0$$

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.

$$y=\frac{-23±\sqrt{23^{2}-4\times 6\left(-18\right)}}{2\times 6}$$

Square 23.

$$y=\frac{-23±\sqrt{529-4\times 6\left(-18\right)}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-23±\sqrt{529-24\left(-18\right)}}{2\times 6}$$

Multiply -24 times -18.

$$y=\frac{-23±\sqrt{529+432}}{2\times 6}$$

Add 529 to 432.

$$y=\frac{-23±\sqrt{961}}{2\times 6}$$

Take the square root of 961.

$$y=\frac{-23±31}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{-23±31}{12}$$

Now solve the equation y=\frac{-23±31}{12} when ± is plus. Add -23 to 31.

$$y=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation y=\frac{-23±31}{12} when ± is minus. Subtract 31 from -23.

$$y=-\frac{54}{12}$$

Reduce the fraction \frac{-54}{12} to lowest terms by extracting and canceling out 6.

$$y=-\frac{9}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{3} for x_{1} and -\frac{9}{2} for x_{2}.

$$6y^{2}+23y-18=6\left(y-\frac{2}{3}\right)\left(y-\left(-\frac{9}{2}\right)\right)$$

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

$$6y^{2}+23y-18=6\left(y-\frac{2}{3}\right)\left(y+\frac{9}{2}\right)$$

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

$$6y^{2}+23y-18=6\times \frac{3y-2}{3}\left(y+\frac{9}{2}\right)$$

Add \frac{9}{2} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6y^{2}+23y-18=6\times \frac{3y-2}{3}\times \frac{2y+9}{2}$$

Multiply \frac{3y-2}{3} times \frac{2y+9}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6y^{2}+23y-18=6\times \frac{\left(3y-2\right)\left(2y+9\right)}{3\times 2}$$

Multiply 3 times 2.

$$6y^{2}+23y-18=6\times \frac{\left(3y-2\right)\left(2y+9\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6y^{2}+23y-18=\left(3y-2\right)\left(2y+9\right)$$

Evaluate

$$\left(3y-2\right)\left(2y+9\right)$$