Solve 24x^2-38x+15

$$24 { x }^{ 2 } -38x+15$$

Factor

$$\left(4x-3\right)\left(6x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 24x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-38 ab=24\times 15=360$$

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 360.

$$-1,-360 -2,-180 -3,-120 -4,-90 -5,-72 -6,-60 -8,-45 -9,-40 -10,-36 -12,-30 -15,-24 -18,-20$$

Calculate the sum for each pair.

$$-1-360=-361 -2-180=-182 -3-120=-123 -4-90=-94 -5-72=-77 -6-60=-66 -8-45=-53 -9-40=-49 -10-36=-46 -12-30=-42 -15-24=-39 -18-20=-38$$

The solution is the pair that gives sum -38.

$$a=-20 b=-18$$

Rewrite 24x^{2}-38x+15 as \left(24x^{2}-20x\right)+\left(-18x+15\right).

$$\left(24x^{2}-20x\right)+\left(-18x+15\right)$$

Factor out 4x in the first and -3 in the second group.

$$4x\left(6x-5\right)-3\left(6x-5\right)$$

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

$$\left(6x-5\right)\left(4x-3\right)$$

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.

$$24x^{2}-38x+15=0$$

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.

$$x=\frac{-\left(-38\right)±\sqrt{\left(-38\right)^{2}-4\times 24\times 15}}{2\times 24}$$

Square -38.

$$x=\frac{-\left(-38\right)±\sqrt{1444-4\times 24\times 15}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-\left(-38\right)±\sqrt{1444-96\times 15}}{2\times 24}$$

Multiply -96 times 15.

$$x=\frac{-\left(-38\right)±\sqrt{1444-1440}}{2\times 24}$$

Add 1444 to -1440.

$$x=\frac{-\left(-38\right)±\sqrt{4}}{2\times 24}$$

Take the square root of 4.

$$x=\frac{-\left(-38\right)±2}{2\times 24}$$

The opposite of -38 is 38.

$$x=\frac{38±2}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{38±2}{48}$$

Now solve the equation x=\frac{38±2}{48} when ± is plus. Add 38 to 2.

$$x=\frac{40}{48}$$

Reduce the fraction \frac{40}{48} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{38±2}{48} when ± is minus. Subtract 2 from 38.

$$x=\frac{36}{48}$$

Reduce the fraction \frac{36}{48} to lowest terms by extracting and canceling out 12.

$$x=\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{6} for x_{1} and \frac{3}{4} for x_{2}.

$$24x^{2}-38x+15=24\left(x-\frac{5}{6}\right)\left(x-\frac{3}{4}\right)$$

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

$$24x^{2}-38x+15=24\times \frac{6x-5}{6}\left(x-\frac{3}{4}\right)$$

Subtract \frac{3}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$24x^{2}-38x+15=24\times \frac{6x-5}{6}\times \frac{4x-3}{4}$$

Multiply \frac{6x-5}{6} times \frac{4x-3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24x^{2}-38x+15=24\times \frac{\left(6x-5\right)\left(4x-3\right)}{6\times 4}$$

Multiply 6 times 4.

$$24x^{2}-38x+15=24\times \frac{\left(6x-5\right)\left(4x-3\right)}{24}$$

Cancel out 24, the greatest common factor in 24 and 24.

$$24x^{2}-38x+15=\left(6x-5\right)\left(4x-3\right)$$

Evaluate

$$\left(4x-3\right)\left(6x-5\right)$$