Solve y^4-8y^3+15y^2=

$$y ^ { 4 } - 8 y ^ { 3 } + 15 y ^ { 2 } =$$

Factor

$$\left(y-5\right)\left(y-3\right)y^{2}$$

Solution Steps

Factor out y^{2}.

$$y^{2}\left(y^{2}-8y+15\right)$$

Consider y^{2}-8y+15. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 15=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite y^{2}-8y+15 as \left(y^{2}-5y\right)+\left(-3y+15\right).

$$\left(y^{2}-5y\right)+\left(-3y+15\right)$$

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

$$y\left(y-5\right)-3\left(y-5\right)$$

Factor out common term y-5 by using distributive property.

$$\left(y-5\right)\left(y-3\right)$$

Rewrite the complete factored expression.

$$y^{2}\left(y-5\right)\left(y-3\right)$$

Evaluate

$$\left(y-5\right)\left(y-3\right)y^{2}$$