Solve 36x^2-12x-15

$$36 { x }^{ 2 } -12x-15$$

Factor

$$3\left(6x-5\right)\left(2x+1\right)$$

Solution Steps

Factor out 3.

$$3\left(12x^{2}-4x-5\right)$$

Consider 12x^{2}-4x-5. Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=12\left(-5\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -4.

$$a=-10 b=6$$

Rewrite 12x^{2}-4x-5 as \left(12x^{2}-10x\right)+\left(6x-5\right).

$$\left(12x^{2}-10x\right)+\left(6x-5\right)$$

Factor out 2x in 12x^{2}-10x.

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

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

$$\left(6x-5\right)\left(2x+1\right)$$

Rewrite the complete factored expression.

$$3\left(6x-5\right)\left(2x+1\right)$$

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.

$$36x^{2}-12x-15=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 36\left(-15\right)}}{2\times 36}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 36\left(-15\right)}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-\left(-12\right)±\sqrt{144-144\left(-15\right)}}{2\times 36}$$

Multiply -144 times -15.

$$x=\frac{-\left(-12\right)±\sqrt{144+2160}}{2\times 36}$$

Add 144 to 2160.

$$x=\frac{-\left(-12\right)±\sqrt{2304}}{2\times 36}$$

Take the square root of 2304.

$$x=\frac{-\left(-12\right)±48}{2\times 36}$$

The opposite of -12 is 12.

$$x=\frac{12±48}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{12±48}{72}$$

Now solve the equation x=\frac{12±48}{72} when ± is plus. Add 12 to 48.

$$x=\frac{60}{72}$$

Reduce the fraction \frac{60}{72} to lowest terms by extracting and canceling out 12.

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

Now solve the equation x=\frac{12±48}{72} when ± is minus. Subtract 48 from 12.

$$x=-\frac{36}{72}$$

Reduce the fraction \frac{-36}{72} to lowest terms by extracting and canceling out 36.

$$x=-\frac{1}{2}$$

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{1}{2} for x_{2}.

$$36x^{2}-12x-15=36\left(x-\frac{5}{6}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$36x^{2}-12x-15=36\left(x-\frac{5}{6}\right)\left(x+\frac{1}{2}\right)$$

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

$$36x^{2}-12x-15=36\times \frac{6x-5}{6}\left(x+\frac{1}{2}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$36x^{2}-12x-15=36\times \frac{6x-5}{6}\times \frac{2x+1}{2}$$

Multiply \frac{6x-5}{6} times \frac{2x+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$36x^{2}-12x-15=36\times \frac{\left(6x-5\right)\left(2x+1\right)}{6\times 2}$$

Multiply 6 times 2.

$$36x^{2}-12x-15=36\times \frac{\left(6x-5\right)\left(2x+1\right)}{12}$$

Cancel out 12, the greatest common factor in 36 and 12.

$$36x^{2}-12x-15=3\left(6x-5\right)\left(2x+1\right)$$

Evaluate

$$36x^{2}-12x-15$$