Solve 9y^2-23y+14

$$9 { y }^{ 2 } -23y+14$$

Factor

$$\left(y-1\right)\left(9y-14\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9y^{2}+ay+by+14. To find a and b, set up a system to be solved.

$$a+b=-23 ab=9\times 14=126$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 126.

$$-1,-126 -2,-63 -3,-42 -6,-21 -7,-18 -9,-14$$

Calculate the sum for each pair.

$$-1-126=-127 -2-63=-65 -3-42=-45 -6-21=-27 -7-18=-25 -9-14=-23$$

The solution is the pair that gives sum -23.

$$a=-14 b=-9$$

Rewrite 9y^{2}-23y+14 as \left(9y^{2}-14y\right)+\left(-9y+14\right).

$$\left(9y^{2}-14y\right)+\left(-9y+14\right)$$

Factor out y in the first and -1 in the second group.

$$y\left(9y-14\right)-\left(9y-14\right)$$

Factor out common term 9y-14 by using distributive property.

$$\left(9y-14\right)\left(y-1\right)$$

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.

$$9y^{2}-23y+14=0$$

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.

$$y=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 9\times 14}}{2\times 9}$$

Square -23.

$$y=\frac{-\left(-23\right)±\sqrt{529-4\times 9\times 14}}{2\times 9}$$

Multiply -4 times 9.

$$y=\frac{-\left(-23\right)±\sqrt{529-36\times 14}}{2\times 9}$$

Multiply -36 times 14.

$$y=\frac{-\left(-23\right)±\sqrt{529-504}}{2\times 9}$$

Add 529 to -504.

$$y=\frac{-\left(-23\right)±\sqrt{25}}{2\times 9}$$

Take the square root of 25.

$$y=\frac{-\left(-23\right)±5}{2\times 9}$$

The opposite of -23 is 23.

$$y=\frac{23±5}{2\times 9}$$

Multiply 2 times 9.

$$y=\frac{23±5}{18}$$

Now solve the equation y=\frac{23±5}{18} when ± is plus. Add 23 to 5.

$$y=\frac{28}{18}$$

Reduce the fraction \frac{28}{18} to lowest terms by extracting and canceling out 2.

$$y=\frac{14}{9}$$

Now solve the equation y=\frac{23±5}{18} when ± is minus. Subtract 5 from 23.

$$y=\frac{18}{18}$$

Divide 18 by 18.

$$y=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{14}{9} for x_{1} and 1 for x_{2}.

$$9y^{2}-23y+14=9\left(y-\frac{14}{9}\right)\left(y-1\right)$$

Subtract \frac{14}{9} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9y^{2}-23y+14=9\times \frac{9y-14}{9}\left(y-1\right)$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9y^{2}-23y+14=\left(9y-14\right)\left(y-1\right)$$

Evaluate

$$\left(y-1\right)\left(9y-14\right)$$