Solve 12x^2+22x-20

$$12 { x }^{ 2 } +22x-20$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(6x^{2}+11x-10\right)$$

Consider 6x^{2}+11x-10. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=11 ab=6\left(-10\right)=-60$$

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 11.

$$a=-4 b=15$$

Rewrite 6x^{2}+11x-10 as \left(6x^{2}-4x\right)+\left(15x-10\right).

$$\left(6x^{2}-4x\right)+\left(15x-10\right)$$

Factor out 2x in the first and 5 in the second group.

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

Factor out common term 3x-2 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$12x^{2}+22x-20=0$$

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.

$$x=\frac{-22±\sqrt{22^{2}-4\times 12\left(-20\right)}}{2\times 12}$$

Square 22.

$$x=\frac{-22±\sqrt{484-4\times 12\left(-20\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-22±\sqrt{484-48\left(-20\right)}}{2\times 12}$$

Multiply -48 times -20.

$$x=\frac{-22±\sqrt{484+960}}{2\times 12}$$

Add 484 to 960.

$$x=\frac{-22±\sqrt{1444}}{2\times 12}$$

Take the square root of 1444.

$$x=\frac{-22±38}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-22±38}{24}$$

Now solve the equation x=\frac{-22±38}{24} when ± is plus. Add -22 to 38.

$$x=\frac{16}{24}$$

Reduce the fraction \frac{16}{24} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{-22±38}{24} when ± is minus. Subtract 38 from -22.

$$x=-\frac{60}{24}$$

Reduce the fraction \frac{-60}{24} to lowest terms by extracting and canceling out 12.

$$x=-\frac{5}{2}$$

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

$$12x^{2}+22x-20=12\left(x-\frac{2}{3}\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$12x^{2}+22x-20=12\left(x-\frac{2}{3}\right)\left(x+\frac{5}{2}\right)$$

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

$$12x^{2}+22x-20=12\times \frac{3x-2}{3}\left(x+\frac{5}{2}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+22x-20=12\times \frac{3x-2}{3}\times \frac{2x+5}{2}$$

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

$$12x^{2}+22x-20=12\times \frac{\left(3x-2\right)\left(2x+5\right)}{3\times 2}$$

Multiply 3 times 2.

$$12x^{2}+22x-20=12\times \frac{\left(3x-2\right)\left(2x+5\right)}{6}$$

Cancel out 6, the greatest common factor in 12 and 6.

$$12x^{2}+22x-20=2\left(3x-2\right)\left(2x+5\right)$$

Evaluate

$$12x^{2}+22x-20$$