Solve x^2+5x-36

$${ x }^{ 2 } +5x-36$$

Factor

$$\left(x-4\right)\left(x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-36. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-36\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=-4 b=9$$

Rewrite x^{2}+5x-36 as \left(x^{2}-4x\right)+\left(9x-36\right).

$$\left(x^{2}-4x\right)+\left(9x-36\right)$$

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

$$x\left(x-4\right)+9\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+9\right)$$

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.

$$x^{2}+5x-36=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-36\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$x=\frac{-5±\sqrt{25+144}}{2}$$

Add 25 to 144.

$$x=\frac{-5±\sqrt{169}}{2}$$

Take the square root of 169.

$$x=\frac{-5±13}{2}$$

Now solve the equation x=\frac{-5±13}{2} when ± is plus. Add -5 to 13.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{-5±13}{2} when ± is minus. Subtract 13 from -5.

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

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 4 for x_{1} and -9 for x_{2}.

$$x^{2}+5x-36=\left(x-4\right)\left(x-\left(-9\right)\right)$$

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

$$x^{2}+5x-36=\left(x-4\right)\left(x+9\right)$$

Evaluate

$$\left(x-4\right)\left(x+9\right)$$