Solve y^4-5y^2-36

$$y ^ { 4 } - 5 y ^ { 2 } - 36$$

Factor

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

Solution Steps

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

$$y^{4}-5y^{2}-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$y=3$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{4}-5y^{2}-36 by y-3 to get y^{3}+3y^{2}+4y+12. To factor the result, solve the equation where it equals to 0.

$$y^{3}+3y^{2}+4y+12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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.

$$y=-3$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{3}+3y^{2}+4y+12 by y+3 to get y^{2}+4. To factor the result, solve the equation where it equals to 0.

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

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 4 for c in the quadratic formula.

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

Do the calculations.

$$y=\frac{0±\sqrt{-16}}{2}$$

Polynomial y^{2}+4 is not factored since it does not have any rational roots.

$$y^{2}+4$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

$$y^{4}-5y^{2}-36$$