Solve y^3-13y^2+36y

$$y ^ { 3 } - 13 y ^ { 2 } + 36 y$$

Factor

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

Solution Steps

Factor out y.

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

Consider y^{2}-13y+36. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+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 y^{2}-13y+36 as \left(y^{2}-9y\right)+\left(-4y+36\right).

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

Factor out y in the first and -4 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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