Solve 2x^2+11x+12

$$2 { x }^{ 2 } +11x+12$$

Factor

$$\left(x+4\right)\left(2x+3\right)$$

Solution Steps

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

$$a+b=11 ab=2\times 12=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=3 b=8$$

Rewrite 2x^{2}+11x+12 as \left(2x^{2}+3x\right)+\left(8x+12\right).

$$\left(2x^{2}+3x\right)+\left(8x+12\right)$$

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

$$x\left(2x+3\right)+4\left(2x+3\right)$$

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

$$\left(2x+3\right)\left(x+4\right)$$

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.

$$2x^{2}+11x+12=0$$

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.

$$x=\frac{-11±\sqrt{11^{2}-4\times 2\times 12}}{2\times 2}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 2\times 12}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-11±\sqrt{121-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$x=\frac{-11±\sqrt{121-96}}{2\times 2}$$

Add 121 to -96.

$$x=\frac{-11±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-11±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-11±5}{4}$$

Now solve the equation x=\frac{-11±5}{4} when ± is plus. Add -11 to 5.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-11±5}{4} when ± is minus. Subtract 5 from -11.

$$x=-\frac{16}{4}$$

Divide -16 by 4.

$$x=-4$$

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

$$2x^{2}+11x+12=2\left(x-\left(-\frac{3}{2}\right)\right)\left(x-\left(-4\right)\right)$$

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

$$2x^{2}+11x+12=2\left(x+\frac{3}{2}\right)\left(x+4\right)$$

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

$$2x^{2}+11x+12=2\times \frac{2x+3}{2}\left(x+4\right)$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2x^{2}+11x+12=\left(2x+3\right)\left(x+4\right)$$

Evaluate

$$\left(x+4\right)\left(2x+3\right)$$