Solve 12y^2-28y+15

$$12 { y }^{ 2 } -28y+15$$

Factor

$$\left(2y-3\right)\left(6y-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12y^{2}+ay+by+15. To find a and b, set up a system to be solved.

$$a+b=-28 ab=12\times 15=180$$

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 180.

$$-1,-180 -2,-90 -3,-60 -4,-45 -5,-36 -6,-30 -9,-20 -10,-18 -12,-15$$

Calculate the sum for each pair.

$$-1-180=-181 -2-90=-92 -3-60=-63 -4-45=-49 -5-36=-41 -6-30=-36 -9-20=-29 -10-18=-28 -12-15=-27$$

The solution is the pair that gives sum -28.

$$a=-18 b=-10$$

Rewrite 12y^{2}-28y+15 as \left(12y^{2}-18y\right)+\left(-10y+15\right).

$$\left(12y^{2}-18y\right)+\left(-10y+15\right)$$

Factor out 6y in the first and -5 in the second group.

$$6y\left(2y-3\right)-5\left(2y-3\right)$$

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

$$\left(2y-3\right)\left(6y-5\right)$$

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.

$$12y^{2}-28y+15=0$$

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.

$$y=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 12\times 15}}{2\times 12}$$

Square -28.

$$y=\frac{-\left(-28\right)±\sqrt{784-4\times 12\times 15}}{2\times 12}$$

Multiply -4 times 12.

$$y=\frac{-\left(-28\right)±\sqrt{784-48\times 15}}{2\times 12}$$

Multiply -48 times 15.

$$y=\frac{-\left(-28\right)±\sqrt{784-720}}{2\times 12}$$

Add 784 to -720.

$$y=\frac{-\left(-28\right)±\sqrt{64}}{2\times 12}$$

Take the square root of 64.

$$y=\frac{-\left(-28\right)±8}{2\times 12}$$

The opposite of -28 is 28.

$$y=\frac{28±8}{2\times 12}$$

Multiply 2 times 12.

$$y=\frac{28±8}{24}$$

Now solve the equation y=\frac{28±8}{24} when ± is plus. Add 28 to 8.

$$y=\frac{36}{24}$$

Reduce the fraction \frac{36}{24} to lowest terms by extracting and canceling out 12.

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

Now solve the equation y=\frac{28±8}{24} when ± is minus. Subtract 8 from 28.

$$y=\frac{20}{24}$$

Reduce the fraction \frac{20}{24} to lowest terms by extracting and canceling out 4.

$$y=\frac{5}{6}$$

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 \frac{5}{6} for x_{2}.

$$12y^{2}-28y+15=12\left(y-\frac{3}{2}\right)\left(y-\frac{5}{6}\right)$$

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

$$12y^{2}-28y+15=12\times \frac{2y-3}{2}\left(y-\frac{5}{6}\right)$$

Subtract \frac{5}{6} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12y^{2}-28y+15=12\times \frac{2y-3}{2}\times \frac{6y-5}{6}$$

Multiply \frac{2y-3}{2} times \frac{6y-5}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12y^{2}-28y+15=12\times \frac{\left(2y-3\right)\left(6y-5\right)}{2\times 6}$$

Multiply 2 times 6.

$$12y^{2}-28y+15=12\times \frac{\left(2y-3\right)\left(6y-5\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12y^{2}-28y+15=\left(2y-3\right)\left(6y-5\right)$$

Evaluate

$$\left(2y-3\right)\left(6y-5\right)$$