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3\left(y^{7}+4y^{6}-5y^{5}\right)
Factor out 3.
y^{5}\left(y^{2}+4y-5\right)
Consider y^{7}+4y^{6}-5y^{5}. Factor out y^{5}.
a+b=4 ab=1\left(-5\right)=-5
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=-1 b=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.
\left(y^{2}-y\right)+\left(5y-5\right)
Rewrite y^{2}+4y-5 as \left(y^{2}-y\right)+\left(5y-5\right).
y\left(y-1\right)+5\left(y-1\right)
Factor out y in the first and 5 in the second group.
\left(y-1\right)\left(y+5\right)
Factor out common term y-1 by using distributive property.
3y^{5}\left(y-1\right)\left(y+5\right)
Rewrite the complete factored expression.