Solve 12x^2-5x-3

$$12 { x }^{ 2 } -5x-3$$

Factor

$$\left(4x-3\right)\left(3x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-5 ab=12\left(-3\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=-9 b=4$$

Rewrite 12x^{2}-5x-3 as \left(12x^{2}-9x\right)+\left(4x-3\right).

$$\left(12x^{2}-9x\right)+\left(4x-3\right)$$

Factor out 3x in 12x^{2}-9x.

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

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(3x+1\right)$$

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.

$$12x^{2}-5x-3=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 12\left(-3\right)}}{2\times 12}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 12\left(-3\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-5\right)±\sqrt{25-48\left(-3\right)}}{2\times 12}$$

Multiply -48 times -3.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\times 12}$$

Add 25 to 144.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\times 12}$$

Take the square root of 169.

$$x=\frac{-\left(-5\right)±13}{2\times 12}$$

The opposite of -5 is 5.

$$x=\frac{5±13}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{5±13}{24}$$

Now solve the equation x=\frac{5±13}{24} when ± is plus. Add 5 to 13.

$$x=\frac{18}{24}$$

Reduce the fraction \frac{18}{24} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{5±13}{24} when ± is minus. Subtract 13 from 5.

$$x=-\frac{8}{24}$$

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

$$x=-\frac{1}{3}$$

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

$$12x^{2}-5x-3=12\left(x-\frac{3}{4}\right)\left(x-\left(-\frac{1}{3}\right)\right)$$

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

$$12x^{2}-5x-3=12\left(x-\frac{3}{4}\right)\left(x+\frac{1}{3}\right)$$

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

$$12x^{2}-5x-3=12\times \frac{4x-3}{4}\left(x+\frac{1}{3}\right)$$

Add \frac{1}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-5x-3=12\times \frac{4x-3}{4}\times \frac{3x+1}{3}$$

Multiply \frac{4x-3}{4} times \frac{3x+1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-5x-3=12\times \frac{\left(4x-3\right)\left(3x+1\right)}{4\times 3}$$

Multiply 4 times 3.

$$12x^{2}-5x-3=12\times \frac{\left(4x-3\right)\left(3x+1\right)}{12}$$

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

$$12x^{2}-5x-3=\left(4x-3\right)\left(3x+1\right)$$

Evaluate

$$\left(4x-3\right)\left(3x+1\right)$$