Solve x^2+11x/2+6=

$$x ^ { 2 } + \frac { 11 x } { 2 } + 6 =$$

Evaluate

$$x^{2}+\frac{11x}{2}+6$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}+6 times \frac{2}{2}.

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

Since \frac{2\left(x^{2}+6\right)}{2} and \frac{11x}{2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2\left(x^{2}+6\right)+11x.

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

Factor

$$\frac{\left(x+4\right)\left(2x+3\right)}{2}$$

Solution Steps

Factor out \frac{1}{2}.

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

Consider 2x^{2}+11x+12. 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.

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

Rewrite the complete factored expression.

$$\frac{\left(2x+3\right)\left(x+4\right)}{2}$$