Solve 15x^2-28x+12

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

Factor

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

Solution Steps

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

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

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

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

Factor out 3x in the first and -2 in the second group.

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

Factor out common term 5x-6 by using distributive property.

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

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.

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

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.

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

Square -28.

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

Multiply -4 times 15.

$$x=\frac{-\left(-28\right)±\sqrt{784-60\times 12}}{2\times 15}$$

Multiply -60 times 12.

$$x=\frac{-\left(-28\right)±\sqrt{784-720}}{2\times 15}$$

Add 784 to -720.

$$x=\frac{-\left(-28\right)±\sqrt{64}}{2\times 15}$$

Take the square root of 64.

$$x=\frac{-\left(-28\right)±8}{2\times 15}$$

The opposite of -28 is 28.

$$x=\frac{28±8}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{28±8}{30}$$

Now solve the equation x=\frac{28±8}{30} when ± is plus. Add 28 to 8.

$$x=\frac{36}{30}$$

Reduce the fraction \frac{36}{30} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{28±8}{30} when ± is minus. Subtract 8 from 28.

$$x=\frac{20}{30}$$

Reduce the fraction \frac{20}{30} to lowest terms by extracting and canceling out 10.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{6}{5} for x_{1} and \frac{2}{3} for x_{2}.

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

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

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

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

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

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

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

Multiply 5 times 3.

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

Cancel out 15, the greatest common factor in 15 and 15.

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 15

$$x ^ 2 -\frac{28}{15}x +\frac{4}{5} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{28}{15} rs = \frac{4}{5}$$

Two numbers r and s sum up to \frac{28}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{28}{15} = \frac{14}{15}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{14}{15} - u s = \frac{14}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{5}

$$(\frac{14}{15} - u) (\frac{14}{15} + u) = \frac{4}{5}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{196}{225} - u^2 = \frac{4}{5}$$

Simplify the expression by subtracting \frac{196}{225} on both sides

$$-u^2 = \frac{4}{5}-\frac{196}{225} = -\frac{16}{225}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{16}{225} u = \pm\sqrt{\frac{16}{225}} = \pm \frac{4}{15}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{14}{15} - \frac{4}{15} = 0.667 s = \frac{14}{15} + \frac{4}{15} = 1.200$$

Evaluate

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