Solve 2y^4+3y^3+3y-2

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

Factor

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

Solution Steps

To factor the expression, solve the equation where it equals to 0.

$$2y^{4}+3y^{3}+3y-2=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±1,±2,±\frac{1}{2}$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$y=-2$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 2y^{4}+3y^{3}+3y-2 by y+2 to get 2y^{3}-y^{2}+2y-1. To factor the result, solve the equation where it equals to 0.

$$2y^{3}-y^{2}+2y-1=0$$

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. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$y=\frac{1}{2}$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 2y^{3}-y^{2}+2y-1 by 2\left(y-\frac{1}{2}\right)=2y-1 to get y^{2}+1. To factor the result, solve the equation where it equals to 0.

$$y^{2}+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 0 for b, and 1 for c in the quadratic formula.

$$y=\frac{0±\sqrt{0^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$y=\frac{0±\sqrt{-4}}{2}$$

Polynomial y^{2}+1 is not factored since it does not have any rational roots.

$$y^{2}+1$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

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