Solve 72x^2-12x-40

$$72 { x }^{ 2 } -12x-40$$

Factor

$$4\left(6x-5\right)\left(3x+2\right)$$

Solution Steps

Factor out 4.

$$4\left(18x^{2}-3x-10\right)$$

Consider 18x^{2}-3x-10. Factor the expression by grouping. First, the expression needs to be rewritten as 18x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=18\left(-10\right)=-180$$

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 -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.

$$1-180=-179 2-90=-88 3-60=-57 4-45=-41 5-36=-31 6-30=-24 9-20=-11 10-18=-8 12-15=-3$$

The solution is the pair that gives sum -3.

$$a=-15 b=12$$

Rewrite 18x^{2}-3x-10 as \left(18x^{2}-15x\right)+\left(12x-10\right).

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

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

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

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

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

Rewrite the complete factored expression.

$$4\left(6x-5\right)\left(3x+2\right)$$

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.

$$72x^{2}-12x-40=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 72\left(-40\right)}}{2\times 72}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 72\left(-40\right)}}{2\times 72}$$

Multiply -4 times 72.

$$x=\frac{-\left(-12\right)±\sqrt{144-288\left(-40\right)}}{2\times 72}$$

Multiply -288 times -40.

$$x=\frac{-\left(-12\right)±\sqrt{144+11520}}{2\times 72}$$

Add 144 to 11520.

$$x=\frac{-\left(-12\right)±\sqrt{11664}}{2\times 72}$$

Take the square root of 11664.

$$x=\frac{-\left(-12\right)±108}{2\times 72}$$

The opposite of -12 is 12.

$$x=\frac{12±108}{2\times 72}$$

Multiply 2 times 72.

$$x=\frac{12±108}{144}$$

Now solve the equation x=\frac{12±108}{144} when ± is plus. Add 12 to 108.

$$x=\frac{120}{144}$$

Reduce the fraction \frac{120}{144} to lowest terms by extracting and canceling out 24.

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

Now solve the equation x=\frac{12±108}{144} when ± is minus. Subtract 108 from 12.

$$x=-\frac{96}{144}$$

Reduce the fraction \frac{-96}{144} to lowest terms by extracting and canceling out 48.

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

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

$$72x^{2}-12x-40=72\left(x-\frac{5}{6}\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$72x^{2}-12x-40=72\left(x-\frac{5}{6}\right)\left(x+\frac{2}{3}\right)$$

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

$$72x^{2}-12x-40=72\times \frac{6x-5}{6}\left(x+\frac{2}{3}\right)$$

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

$$72x^{2}-12x-40=72\times \frac{6x-5}{6}\times \frac{3x+2}{3}$$

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

$$72x^{2}-12x-40=72\times \frac{\left(6x-5\right)\left(3x+2\right)}{6\times 3}$$

Multiply 6 times 3.

$$72x^{2}-12x-40=72\times \frac{\left(6x-5\right)\left(3x+2\right)}{18}$$

Cancel out 18, the greatest common factor in 72 and 18.

$$72x^{2}-12x-40=4\left(6x-5\right)\left(3x+2\right)$$

Evaluate

$$72x^{2}-12x-40$$