Solve 2y^7+8y^6-10y^5

$$2 { y }^{ 7 } +8 { y }^{ 6 } -10 { y }^{ 5 }$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(y^{7}+4y^{6}-5y^{5}\right)$$

Consider y^{7}+4y^{6}-5y^{5}. Factor out y^{5}.

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

Consider y^{2}+4y-5. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-5. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-5\right)=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=5$$

Rewrite y^{2}+4y-5 as \left(y^{2}-y\right)+\left(5y-5\right).

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

Factor out y in the first and 5 in the second group.

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

Factor out common term y-1 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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