Solve f(y)=16y^4-8y+32y^3-40y^2=

$$f ( y ) = 16 y ^ { 4 } - 8 y + 32 y ^ { 3 } - 40 y ^ { 2 } =$$

Factor

$$8y\left(y-1\right)\left(2y^{2}+6y+1\right)$$

Solution Steps

Factor out 8.

$$8\left(2y^{4}-y+4y^{3}-5y^{2}\right)$$

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

$$y\left(2y^{3}-1+4y^{2}-5y\right)$$

Consider 2y^{3}-1+4y^{2}-5y. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 2. One such root is 1. Factor the polynomial by dividing it by y-1.

$$\left(y-1\right)\left(2y^{2}+6y+1\right)$$

Rewrite the complete factored expression. Polynomial 2y^{2}+6y+1 is not factored since it does not have any rational roots.

$$8y\left(y-1\right)\left(2y^{2}+6y+1\right)$$

Evaluate

$$8y\left(y-1\right)\left(2y^{2}+6y+1\right)$$