Solve x^3-13x^2+36x=

$$x ^ { 3 } - 13 x ^ { 2 } + 36 x =$$

Factor

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

Solution Steps

Factor out x.

$$x\left(x^{2}-13x+36\right)$$

Consider x^{2}-13x+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.

$$a+b=-13 ab=1\times 36=36$$

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 -13.

$$a=-9 b=-4$$

Rewrite x^{2}-13x+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.

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

Rewrite the complete factored expression.

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

Evaluate

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