Solve y^4-3y^2-4

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

Factor

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

Solution Steps

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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.

$$y=2$$

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

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

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

$$±2,±1$$

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.

$$y=-2$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{3}+2y^{2}+y+2 by y+2 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.

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

Evaluate

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