Solve y^4-6y^2+8

$${ y }^{ 4 } -6 { y }^{ 2 } +8$$

Factor

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

Solution Steps

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

$$y^{4}-6y^{2}+8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

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}-6y^{2}+8 by y-2 to get y^{3}+2y^{2}-2y-4. To factor the result, solve the equation where it equals to 0.

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

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^{3}+2y^{2}-2y-4 by y+2 to get y^{2}-2. To factor the result, solve the equation where it equals to 0.

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

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 -2 for c in the quadratic formula.

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

Do the calculations.

$$y=\frac{0±2\sqrt{2}}{2}$$

Solve the equation y^{2}-2=0 when ± is plus and when ± is minus.

$$y=-\sqrt{2} y=\sqrt{2}$$

Rewrite the factored expression using the obtained roots. Polynomial y^{2}-2 is not factored since it does not have any rational roots.

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

Evaluate

$$y^{4}-6y^{2}+8$$