Solve 36+x^2-15x

$$36 + x ^ { 2 } - 15 x$$

Factor

$$\left(x-12\right)\left(x-3\right)$$

Solution Steps

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

$$x^{2}-15x+36$$

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.

$$a+b=-15 ab=1\times 36=36$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both 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.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -15.

$$a=-12 b=-3$$

Rewrite x^{2}-15x+36 as \left(x^{2}-12x\right)+\left(-3x+36\right).

$$\left(x^{2}-12x\right)+\left(-3x+36\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-12\right)-3\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(x-3\right)$$

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.

$$x^{2}-15x+36=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 36}}{2}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 36}}{2}$$

Multiply -4 times 36.

$$x=\frac{-\left(-15\right)±\sqrt{225-144}}{2}$$

Add 225 to -144.

$$x=\frac{-\left(-15\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-\left(-15\right)±9}{2}$$

The opposite of -15 is 15.

$$x=\frac{15±9}{2}$$

Now solve the equation x=\frac{15±9}{2} when ± is plus. Add 15 to 9.

$$x=\frac{24}{2}$$

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{15±9}{2} when ± is minus. Subtract 9 from 15.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 12 for x_{1} and 3 for x_{2}.

$$x^{2}-15x+36=\left(x-12\right)\left(x-3\right)$$

Evaluate

$$\left(x-12\right)\left(x-3\right)$$