Solve 14x+x^2+45

$$14 x + x ^ { 2 } + 45$$

Factor

$$\left(x+5\right)\left(x+9\right)$$

Solution Steps

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

$$x^{2}+14x+45$$

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=14 ab=1\times 45=45$$

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 45.

$$1,45 3,15 5,9$$

Calculate the sum for each pair.

$$1+45=46 3+15=18 5+9=14$$

The solution is the pair that gives sum 14.

$$a=5 b=9$$

Rewrite x^{2}+14x+45 as \left(x^{2}+5x\right)+\left(9x+45\right).

$$\left(x^{2}+5x\right)+\left(9x+45\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x+5\right)+9\left(x+5\right)$$

Factor out common term x+5 by using distributive property.

$$\left(x+5\right)\left(x+9\right)$$

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.

$$x^{2}+14x+45=0$$

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.

$$x=\frac{-14±\sqrt{14^{2}-4\times 45}}{2}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 45}}{2}$$

Multiply -4 times 45.

$$x=\frac{-14±\sqrt{196-180}}{2}$$

Add 196 to -180.

$$x=\frac{-14±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-14±4}{2}$$

Now solve the equation x=\frac{-14±4}{2} when ± is plus. Add -14 to 4.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

Now solve the equation x=\frac{-14±4}{2} when ± is minus. Subtract 4 from -14.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and -9 for x_{2}.

$$x^{2}+14x+45=\left(x-\left(-5\right)\right)\left(x-\left(-9\right)\right)$$

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

$$x^{2}+14x+45=\left(x+5\right)\left(x+9\right)$$

Evaluate

$$\left(x+5\right)\left(x+9\right)$$