Solve -36-5x+x^2

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

Factor

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

Solution Steps

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

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

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-9 b=4$$

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

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

Factor out x in the first and 4 in the second group.

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

Factor out common term x-9 by using distributive property.

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

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.

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

Square -5.

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

Multiply -4 times -36.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2}$$

Add 25 to 144.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2}$$

Take the square root of 169.

$$x=\frac{-\left(-5\right)±13}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±13}{2}$$

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

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

Divide 18 by 2.

$$x=9$$

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

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

Divide -8 by 2.

$$x=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 9 for x_{1} and -4 for x_{2}.

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

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

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

Evaluate

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