Solve 2x^2+7x+6

$$2 { x }^{ 2 } +7x+6$$

Factor

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

Solution Steps

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

$$a+b=7 ab=2\times 6=12$$

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 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite 2x^{2}+7x+6 as \left(2x^{2}+3x\right)+\left(4x+6\right).

$$\left(2x^{2}+3x\right)+\left(4x+6\right)$$

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

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

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

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

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.

$$2x^{2}+7x+6=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 6}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-7±\sqrt{49-48}}{2\times 2}$$

Add 49 to -48.

$$x=\frac{-7±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-7±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±1}{4}$$

Now solve the equation x=\frac{-7±1}{4} when ± is plus. Add -7 to 1.

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

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{-7±1}{4} when ± is minus. Subtract 1 from -7.

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

Divide -8 by 4.

$$x=-2$$

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 -2 for x_{2}.

$$2x^{2}+7x+6=2\left(x-\left(-\frac{3}{2}\right)\right)\left(x-\left(-2\right)\right)$$

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

$$2x^{2}+7x+6=2\left(x+\frac{3}{2}\right)\left(x+2\right)$$

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

$$2x^{2}+7x+6=2\times \frac{2x+3}{2}\left(x+2\right)$$

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

$$2x^{2}+7x+6=\left(2x+3\right)\left(x+2\right)$$

Evaluate

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