Solve x^2+13x-90

$${ x }^{ 2 } +13x-90$$

Factor

$$\left(x-5\right)\left(x+18\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-90. To find a and b, set up a system to be solved.

$$a+b=13 ab=1\left(-90\right)=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -90.

$$-1,90 -2,45 -3,30 -5,18 -6,15 -9,10$$

Calculate the sum for each pair.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

The solution is the pair that gives sum 13.

$$a=-5 b=18$$

Rewrite x^{2}+13x-90 as \left(x^{2}-5x\right)+\left(18x-90\right).

$$\left(x^{2}-5x\right)+\left(18x-90\right)$$

Factor out x in the first and 18 in the second group.

$$x\left(x-5\right)+18\left(x-5\right)$$

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

$$\left(x-5\right)\left(x+18\right)$$

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.

$$x^{2}+13x-90=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-90\right)}}{2}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-90\right)}}{2}$$

Multiply -4 times -90.

$$x=\frac{-13±\sqrt{169+360}}{2}$$

Add 169 to 360.

$$x=\frac{-13±\sqrt{529}}{2}$$

Take the square root of 529.

$$x=\frac{-13±23}{2}$$

Now solve the equation x=\frac{-13±23}{2} when ± is plus. Add -13 to 23.

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

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{-13±23}{2} when ± is minus. Subtract 23 from -13.

$$x=-\frac{36}{2}$$

Divide -36 by 2.

$$x=-18$$

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 -18 for x_{2}.

$$x^{2}+13x-90=\left(x-5\right)\left(x-\left(-18\right)\right)$$

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

$$x^{2}+13x-90=\left(x-5\right)\left(x+18\right)$$

Evaluate

$$\left(x-5\right)\left(x+18\right)$$