Solve 20x^2-13x-15

$$20 { x }^{ 2 } -13x-15$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=-13 ab=20\left(-15\right)=-300$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -300.

$$1,-300 2,-150 3,-100 4,-75 5,-60 6,-50 10,-30 12,-25 15,-20$$

Calculate the sum for each pair.

$$1-300=-299 2-150=-148 3-100=-97 4-75=-71 5-60=-55 6-50=-44 10-30=-20 12-25=-13 15-20=-5$$

The solution is the pair that gives sum -13.

$$a=-25 b=12$$

Rewrite 20x^{2}-13x-15 as \left(20x^{2}-25x\right)+\left(12x-15\right).

$$\left(20x^{2}-25x\right)+\left(12x-15\right)$$

Factor out 5x in the first and 3 in the second group.

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

Factor out common term 4x-5 by using distributive property.

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

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.

$$20x^{2}-13x-15=0$$

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.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 20\left(-15\right)}}{2\times 20}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 20\left(-15\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-13\right)±\sqrt{169-80\left(-15\right)}}{2\times 20}$$

Multiply -80 times -15.

$$x=\frac{-\left(-13\right)±\sqrt{169+1200}}{2\times 20}$$

Add 169 to 1200.

$$x=\frac{-\left(-13\right)±\sqrt{1369}}{2\times 20}$$

Take the square root of 1369.

$$x=\frac{-\left(-13\right)±37}{2\times 20}$$

The opposite of -13 is 13.

$$x=\frac{13±37}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{13±37}{40}$$

Now solve the equation x=\frac{13±37}{40} when ± is plus. Add 13 to 37.

$$x=\frac{50}{40}$$

Reduce the fraction \frac{50}{40} to lowest terms by extracting and canceling out 10.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{13±37}{40} when ± is minus. Subtract 37 from 13.

$$x=-\frac{24}{40}$$

Reduce the fraction \frac{-24}{40} to lowest terms by extracting and canceling out 8.

$$x=-\frac{3}{5}$$

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

$$20x^{2}-13x-15=20\left(x-\frac{5}{4}\right)\left(x-\left(-\frac{3}{5}\right)\right)$$

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

$$20x^{2}-13x-15=20\left(x-\frac{5}{4}\right)\left(x+\frac{3}{5}\right)$$

Subtract \frac{5}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-13x-15=20\times \frac{4x-5}{4}\left(x+\frac{3}{5}\right)$$

Add \frac{3}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-13x-15=20\times \frac{4x-5}{4}\times \frac{5x+3}{5}$$

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

$$20x^{2}-13x-15=20\times \frac{\left(4x-5\right)\left(5x+3\right)}{4\times 5}$$

Multiply 4 times 5.

$$20x^{2}-13x-15=20\times \frac{\left(4x-5\right)\left(5x+3\right)}{20}$$

Cancel out 20, the greatest common factor in 20 and 20.

$$20x^{2}-13x-15=\left(4x-5\right)\left(5x+3\right)$$

Evaluate

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