Solve 4y^5-22y^4+10y^3

$$4 y ^ { 5 } - 22 y ^ { 4 } + 10 y ^ { 3 }$$

Factor

$$2\left(y-5\right)\left(2y-1\right)y^{3}$$

Solution Steps

Factor out 2.

$$2\left(2y^{5}-11y^{4}+5y^{3}\right)$$

Consider 2y^{5}-11y^{4}+5y^{3}. Factor out y^{3}.

$$y^{3}\left(2y^{2}-11y+5\right)$$

Consider 2y^{2}-11y+5. Factor the expression by grouping. First, the expression needs to be rewritten as 2y^{2}+ay+by+5. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\times 5=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -11.

$$a=-10 b=-1$$

Rewrite 2y^{2}-11y+5 as \left(2y^{2}-10y\right)+\left(-y+5\right).

$$\left(2y^{2}-10y\right)+\left(-y+5\right)$$

Factor out 2y in the first and -1 in the second group.

$$2y\left(y-5\right)-\left(y-5\right)$$

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

$$\left(y-5\right)\left(2y-1\right)$$

Rewrite the complete factored expression.

$$2y^{3}\left(y-5\right)\left(2y-1\right)$$

Evaluate

$$2\left(y-5\right)\left(2y-1\right)y^{3}$$