Solve 24x^2+17x-20

$$24 { x }^{ 2 } +17x-20$$

Factor

$$\left(8x-5\right)\left(3x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 24x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=17 ab=24\left(-20\right)=-480$$

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 -480.

$$-1,480 -2,240 -3,160 -4,120 -5,96 -6,80 -8,60 -10,48 -12,40 -15,32 -16,30 -20,24$$

Calculate the sum for each pair.

$$-1+480=479 -2+240=238 -3+160=157 -4+120=116 -5+96=91 -6+80=74 -8+60=52 -10+48=38 -12+40=28 -15+32=17 -16+30=14 -20+24=4$$

The solution is the pair that gives sum 17.

$$a=-15 b=32$$

Rewrite 24x^{2}+17x-20 as \left(24x^{2}-15x\right)+\left(32x-20\right).

$$\left(24x^{2}-15x\right)+\left(32x-20\right)$$

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

$$3x\left(8x-5\right)+4\left(8x-5\right)$$

Factor out common term 8x-5 by using distributive property.

$$\left(8x-5\right)\left(3x+4\right)$$

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.

$$24x^{2}+17x-20=0$$

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.

$$x=\frac{-17±\sqrt{17^{2}-4\times 24\left(-20\right)}}{2\times 24}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 24\left(-20\right)}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-17±\sqrt{289-96\left(-20\right)}}{2\times 24}$$

Multiply -96 times -20.

$$x=\frac{-17±\sqrt{289+1920}}{2\times 24}$$

Add 289 to 1920.

$$x=\frac{-17±\sqrt{2209}}{2\times 24}$$

Take the square root of 2209.

$$x=\frac{-17±47}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{-17±47}{48}$$

Now solve the equation x=\frac{-17±47}{48} when ± is plus. Add -17 to 47.

$$x=\frac{30}{48}$$

Reduce the fraction \frac{30}{48} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{8}$$

Now solve the equation x=\frac{-17±47}{48} when ± is minus. Subtract 47 from -17.

$$x=-\frac{64}{48}$$

Reduce the fraction \frac{-64}{48} to lowest terms by extracting and canceling out 16.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{8} for x_{1} and -\frac{4}{3} for x_{2}.

$$24x^{2}+17x-20=24\left(x-\frac{5}{8}\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$24x^{2}+17x-20=24\left(x-\frac{5}{8}\right)\left(x+\frac{4}{3}\right)$$

Subtract \frac{5}{8} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+17x-20=24\times \frac{8x-5}{8}\left(x+\frac{4}{3}\right)$$

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

$$24x^{2}+17x-20=24\times \frac{8x-5}{8}\times \frac{3x+4}{3}$$

Multiply \frac{8x-5}{8} times \frac{3x+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+17x-20=24\times \frac{\left(8x-5\right)\left(3x+4\right)}{8\times 3}$$

Multiply 8 times 3.

$$24x^{2}+17x-20=24\times \frac{\left(8x-5\right)\left(3x+4\right)}{24}$$

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

$$24x^{2}+17x-20=\left(8x-5\right)\left(3x+4\right)$$

Evaluate

$$\left(8x-5\right)\left(3x+4\right)$$