Solve 12x^2+5x-3

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

Factor

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

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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-4 b=9$$

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

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

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

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

Factor out common term 3x-1 by using distributive property.

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

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.

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

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.

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

Square 5.

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

Multiply -4 times 12.

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

Multiply -48 times -3.

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

Add 25 to 144.

$$x=\frac{-5±\sqrt{169}}{2\times 12}$$

Take the square root of 169.

$$x=\frac{-5±13}{2\times 12}$$

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply 3 times 4.

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

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

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

Evaluate

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