Solve 80x^2+23x-15

$$80 { x }^{ 2 } +23x-15$$

Factor

$$\left(16x-5\right)\left(5x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 80x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=23 ab=80\left(-15\right)=-1200$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -1200.

$$-1,1200 -2,600 -3,400 -4,300 -5,240 -6,200 -8,150 -10,120 -12,100 -15,80 -16,75 -20,60 -24,50 -25,48 -30,40$$

Calculate the sum for each pair.

$$-1+1200=1199 -2+600=598 -3+400=397 -4+300=296 -5+240=235 -6+200=194 -8+150=142 -10+120=110 -12+100=88 -15+80=65 -16+75=59 -20+60=40 -24+50=26 -25+48=23 -30+40=10$$

The solution is the pair that gives sum 23.

$$a=-25 b=48$$

Rewrite 80x^{2}+23x-15 as \left(80x^{2}-25x\right)+\left(48x-15\right).

$$\left(80x^{2}-25x\right)+\left(48x-15\right)$$

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

$$5x\left(16x-5\right)+3\left(16x-5\right)$$

Factor out common term 16x-5 by using distributive property.

$$\left(16x-5\right)\left(5x+3\right)$$

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.

$$80x^{2}+23x-15=0$$

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.

$$x=\frac{-23±\sqrt{23^{2}-4\times 80\left(-15\right)}}{2\times 80}$$

Square 23.

$$x=\frac{-23±\sqrt{529-4\times 80\left(-15\right)}}{2\times 80}$$

Multiply -4 times 80.

$$x=\frac{-23±\sqrt{529-320\left(-15\right)}}{2\times 80}$$

Multiply -320 times -15.

$$x=\frac{-23±\sqrt{529+4800}}{2\times 80}$$

Add 529 to 4800.

$$x=\frac{-23±\sqrt{5329}}{2\times 80}$$

Take the square root of 5329.

$$x=\frac{-23±73}{2\times 80}$$

Multiply 2 times 80.

$$x=\frac{-23±73}{160}$$

Now solve the equation x=\frac{-23±73}{160} when ± is plus. Add -23 to 73.

$$x=\frac{50}{160}$$

Reduce the fraction \frac{50}{160} to lowest terms by extracting and canceling out 10.

$$x=\frac{5}{16}$$

Now solve the equation x=\frac{-23±73}{160} when ± is minus. Subtract 73 from -23.

$$x=-\frac{96}{160}$$

Reduce the fraction \frac{-96}{160} to lowest terms by extracting and canceling out 32.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{16} for x_{1} and -\frac{3}{5} for x_{2}.

$$80x^{2}+23x-15=80\left(x-\frac{5}{16}\right)\left(x-\left(-\frac{3}{5}\right)\right)$$

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

$$80x^{2}+23x-15=80\left(x-\frac{5}{16}\right)\left(x+\frac{3}{5}\right)$$

Subtract \frac{5}{16} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$80x^{2}+23x-15=80\times \frac{16x-5}{16}\left(x+\frac{3}{5}\right)$$

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

$$80x^{2}+23x-15=80\times \frac{16x-5}{16}\times \frac{5x+3}{5}$$

Multiply \frac{16x-5}{16} times \frac{5x+3}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$80x^{2}+23x-15=80\times \frac{\left(16x-5\right)\left(5x+3\right)}{16\times 5}$$

Multiply 16 times 5.

$$80x^{2}+23x-15=80\times \frac{\left(16x-5\right)\left(5x+3\right)}{80}$$

Cancel out 80, the greatest common factor in 80 and 80.

$$80x^{2}+23x-15=\left(16x-5\right)\left(5x+3\right)$$

Evaluate

$$\left(16x-5\right)\left(5x+3\right)$$