Solve x^2+5x+6-0

$$x ^ { 2 } + 5 x + 6 - 0$$

Evaluate

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

Solution Steps

Multiply -1 and 0 to get 0.

$$x^{2}+5x+6+0$$

Add 6 and 0 to get 6.

$$x^{2}+5x+6$$

Factor

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

Solution Steps

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.

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

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.

$$x^{2}+5x+6=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 6}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 6}}{2}$$

Multiply -4 times 6.

$$x=\frac{-5±\sqrt{25-24}}{2}$$

Add 25 to -24.

$$x=\frac{-5±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-5±1}{2}$$

Now solve the equation x=\frac{-5±1}{2} when ± is plus. Add -5 to 1.

$$x=-\frac{4}{2}$$

Divide -4 by 2.

$$x=-2$$

Now solve the equation x=\frac{-5±1}{2} when ± is minus. Subtract 1 from -5.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

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}.

$$x^{2}+5x+6=\left(x-\left(-2\right)\right)\left(x-\left(-3\right)\right)$$

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

$$x^{2}+5x+6=\left(x+2\right)\left(x+3\right)$$