Solve x^2+13x+40

$${ x }^{ 2 } +13x+40$$

Factor

$$\left(x+5\right)\left(x+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+40. To find a and b, set up a system to be solved.

$$a+b=13 ab=1\times 40=40$$

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 40.

$$1,40 2,20 4,10 5,8$$

Calculate the sum for each pair.

$$1+40=41 2+20=22 4+10=14 5+8=13$$

The solution is the pair that gives sum 13.

$$a=5 b=8$$

Rewrite x^{2}+13x+40 as \left(x^{2}+5x\right)+\left(8x+40\right).

$$\left(x^{2}+5x\right)+\left(8x+40\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x+5\right)+8\left(x+5\right)$$

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

$$\left(x+5\right)\left(x+8\right)$$

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.

$$x^{2}+13x+40=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\times 40}}{2}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 40}}{2}$$

Multiply -4 times 40.

$$x=\frac{-13±\sqrt{169-160}}{2}$$

Add 169 to -160.

$$x=\frac{-13±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-13±3}{2}$$

Now solve the equation x=\frac{-13±3}{2} when ± is plus. Add -13 to 3.

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

Divide -10 by 2.

$$x=-5$$

Now solve the equation x=\frac{-13±3}{2} when ± is minus. Subtract 3 from -13.

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

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 -8 for x_{2}.

$$x^{2}+13x+40=\left(x-\left(-5\right)\right)\left(x-\left(-8\right)\right)$$

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

$$x^{2}+13x+40=\left(x+5\right)\left(x+8\right)$$

Evaluate

$$\left(x+5\right)\left(x+8\right)$$