Solve 20x^2+38x+12

$$20 { x }^{ 2 } +38x+12$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(10x^{2}+19x+6\right)$$

Consider 10x^{2}+19x+6. Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=19 ab=10\times 6=60$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both 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.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 19.

$$a=4 b=15$$

Rewrite 10x^{2}+19x+6 as \left(10x^{2}+4x\right)+\left(15x+6\right).

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

Factor out 2x in the first and 3 in the second group.

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

Factor out common term 5x+2 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$20x^{2}+38x+12=0$$

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.

$$x=\frac{-38±\sqrt{38^{2}-4\times 20\times 12}}{2\times 20}$$

Square 38.

$$x=\frac{-38±\sqrt{1444-4\times 20\times 12}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-38±\sqrt{1444-80\times 12}}{2\times 20}$$

Multiply -80 times 12.

$$x=\frac{-38±\sqrt{1444-960}}{2\times 20}$$

Add 1444 to -960.

$$x=\frac{-38±\sqrt{484}}{2\times 20}$$

Take the square root of 484.

$$x=\frac{-38±22}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-38±22}{40}$$

Now solve the equation x=\frac{-38±22}{40} when ± is plus. Add -38 to 22.

$$x=-\frac{16}{40}$$

Reduce the fraction \frac{-16}{40} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{-38±22}{40} when ± is minus. Subtract 22 from -38.

$$x=-\frac{60}{40}$$

Reduce the fraction \frac{-60}{40} to lowest terms by extracting and canceling out 20.

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

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

$$20x^{2}+38x+12=20\left(x-\left(-\frac{2}{5}\right)\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$20x^{2}+38x+12=20\left(x+\frac{2}{5}\right)\left(x+\frac{3}{2}\right)$$

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

$$20x^{2}+38x+12=20\times \frac{5x+2}{5}\left(x+\frac{3}{2}\right)$$

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

$$20x^{2}+38x+12=20\times \frac{5x+2}{5}\times \frac{2x+3}{2}$$

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

$$20x^{2}+38x+12=20\times \frac{\left(5x+2\right)\left(2x+3\right)}{5\times 2}$$

Multiply 5 times 2.

$$20x^{2}+38x+12=20\times \frac{\left(5x+2\right)\left(2x+3\right)}{10}$$

Cancel out 10, the greatest common factor in 20 and 10.

$$20x^{2}+38x+12=2\left(5x+2\right)\left(2x+3\right)$$

Evaluate

$$20x^{2}+38x+12$$