Solve 2x^2+12x+10

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

Factor

$$2\left(x+1\right)\left(x+5\right)$$

Solution Steps

Factor out 2.

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

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

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

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=5$$

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

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

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

$$x\left(x+1\right)+5\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+5\right)$$

Rewrite the complete factored expression.

$$2\left(x+1\right)\left(x+5\right)$$

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.

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

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.

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

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 2\times 10}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-12±\sqrt{144-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

$$x=\frac{-12±\sqrt{144-80}}{2\times 2}$$

Add 144 to -80.

$$x=\frac{-12±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$x=\frac{-12±8}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-12±8}{4}$$

Now solve the equation x=\frac{-12±8}{4} when ± is plus. Add -12 to 8.

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

Divide -4 by 4.

$$x=-1$$

Now solve the equation x=\frac{-12±8}{4} when ± is minus. Subtract 8 from -12.

$$x=-\frac{20}{4}$$

Divide -20 by 4.

$$x=-5$$

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

$$2x^{2}+12x+10=2\left(x-\left(-1\right)\right)\left(x-\left(-5\right)\right)$$

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

$$2x^{2}+12x+10=2\left(x+1\right)\left(x+5\right)$$

Evaluate

$$2\left(x+1\right)\left(x+5\right)$$