Solve 2x^2+10x+12

$$2 { x }^{ 2 } +10x+12$$

Factor

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

Solution Steps

Factor out 2.

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

Consider x^{2}+5x+6. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\times 6=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=2 b=3$$

Rewrite x^{2}+5x+6 as \left(x^{2}+2x\right)+\left(3x+6\right).

$$\left(x^{2}+2x\right)+\left(3x+6\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$2x^{2}+10x+12=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\times 12}}{2\times 2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2\times 12}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$x=\frac{-10±\sqrt{100-96}}{2\times 2}$$

Add 100 to -96.

$$x=\frac{-10±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

$$x=\frac{-10±2}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-10±2}{4}$$

Now solve the equation x=\frac{-10±2}{4} when ± is plus. Add -10 to 2.

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

Now solve the equation x=\frac{-10±2}{4} when ± is minus. Subtract 2 from -10.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

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

$$2x^{2}+10x+12=2\left(x-\left(-2\right)\right)\left(x-\left(-3\right)\right)$$

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

$$2x^{2}+10x+12=2\left(x+2\right)\left(x+3\right)$$

Evaluate

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