Solve x^2+33+14x

$$x ^ { 2 } + 33 + 14 x$$

Factor

$$\left(x+3\right)\left(x+11\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+14x+33$$

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+33. To find a and b, set up a system to be solved.

$$a+b=14 ab=1\times 33=33$$

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 33.

$$1,33 3,11$$

Calculate the sum for each pair.

$$1+33=34 3+11=14$$

The solution is the pair that gives sum 14.

$$a=3 b=11$$

Rewrite x^{2}+14x+33 as \left(x^{2}+3x\right)+\left(11x+33\right).

$$\left(x^{2}+3x\right)+\left(11x+33\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(x+3\right)+11\left(x+3\right)$$

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

$$\left(x+3\right)\left(x+11\right)$$

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.

$$x^{2}+14x+33=0$$

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.

$$x=\frac{-14±\sqrt{14^{2}-4\times 33}}{2}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 33}}{2}$$

Multiply -4 times 33.

$$x=\frac{-14±\sqrt{196-132}}{2}$$

Add 196 to -132.

$$x=\frac{-14±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-14±8}{2}$$

Now solve the equation x=\frac{-14±8}{2} when ± is plus. Add -14 to 8.

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

Divide -6 by 2.

$$x=-3$$

Now solve the equation x=\frac{-14±8}{2} when ± is minus. Subtract 8 from -14.

$$x=-\frac{22}{2}$$

Divide -22 by 2.

$$x=-11$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3 for x_{1} and -11 for x_{2}.

$$x^{2}+14x+33=\left(x-\left(-3\right)\right)\left(x-\left(-11\right)\right)$$

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

$$x^{2}+14x+33=\left(x+3\right)\left(x+11\right)$$

Evaluate

$$\left(x+3\right)\left(x+11\right)$$