Solve 16x^4-8x^2+1

$$16 x ^ { 4 } - 8 x ^ { 2 } + 1$$

Factor

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

Solution Steps

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

$$16x^{4}-8x^{2}+1=0$$

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

$$±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 16x^{4}-8x^{2}+1 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 8x^{3}+4x^{2}-2x-1. To factor the result, solve the equation where it equals to 0.

$$8x^{3}+4x^{2}-2x-1=0$$

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

$$±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}+4x^{2}-2x-1 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 4x^{2}+4x+1. To factor the result, solve the equation where it equals to 0.

$$4x^{2}+4x+1=0$$

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 4 for a, 4 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-4±0}{8}$$

Solutions are the same.

$$x=-\frac{1}{2}$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

$$\left(4x^{2}-1\right)^{2}$$