Solve 5x^2+4x-12=

$$5 { x }^{ 2 } +4x-12=$$

Factor

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

Solution Steps

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

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

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 -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.

$$a=-6 b=10$$

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

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

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

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

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

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

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.

$$5x^{2}+4x-12=0$$

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.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 5\left(-12\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-4±\sqrt{16-20\left(-12\right)}}{2\times 5}$$

Multiply -20 times -12.

$$x=\frac{-4±\sqrt{16+240}}{2\times 5}$$

Add 16 to 240.

$$x=\frac{-4±\sqrt{256}}{2\times 5}$$

Take the square root of 256.

$$x=\frac{-4±16}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-4±16}{10}$$

Now solve the equation x=\frac{-4±16}{10} when ± is plus. Add -4 to 16.

$$x=\frac{12}{10}$$

Reduce the fraction \frac{12}{10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-4±16}{10} when ± is minus. Subtract 16 from -4.

$$x=-\frac{20}{10}$$

Divide -20 by 10.

$$x=-2$$

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

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

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

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

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

$$5x^{2}+4x-12=5\times \frac{5x-6}{5}\left(x+2\right)$$

Cancel out 5, the greatest common factor in 5 and 5.

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

Evaluate

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