Solve -3y+7y+4y^3-2y^2+6y^3-7y^3-5y

$$- 3 y + 7 y + 4 y ^ { 3 } - 2 y ^ { 2 } + 6 y ^ { 3 } - 7 y ^ { 3 } - 5 y$$

Evaluate

$$y\left(y-1\right)\left(3y+1\right)$$

Solution Steps

Combine -3y and 7y to get 4y.

$$4y+4y^{3}-2y^{2}+6y^{3}-7y^{3}-5y$$

Combine 4y^{3} and 6y^{3} to get 10y^{3}.

$$4y+10y^{3}-2y^{2}-7y^{3}-5y$$

Combine 10y^{3} and -7y^{3} to get 3y^{3}.

$$4y+3y^{3}-2y^{2}-5y$$

Combine 4y and -5y to get -y.

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

Factor

$$y\left(y-1\right)\left(3y+1\right)$$

Solution Steps

Factor out y.

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

Consider -3+7+4y^{2}-2y+6y^{2}-7y^{2}-5. Multiply and combine like terms.

$$3y^{2}-2y-1$$

Consider 3y^{2}-2y-1. Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by-1. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-1\right)=-3$$

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

$$a=-3 b=1$$

Rewrite 3y^{2}-2y-1 as \left(3y^{2}-3y\right)+\left(y-1\right).

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

Factor out 3y in 3y^{2}-3y.

$$3y\left(y-1\right)+y-1$$

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

$$\left(y-1\right)\left(3y+1\right)$$

Rewrite the complete factored expression.

$$y\left(y-1\right)\left(3y+1\right)$$