Solve 9x^4-10x^2+1

$$9 x ^ { 4 } - 10 x ^ { 2 } + 1$$

Factor

$$\left(x-1\right)\left(3x-1\right)\left(x+1\right)\left(3x+1\right)$$

Solution Steps

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

$$9x^{4}-10x^{2}+1=0$$

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

$$±\frac{1}{9},±\frac{1}{3},±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 9x^{4}-10x^{2}+1 by x-1 to get 9x^{3}+9x^{2}-x-1. To factor the result, solve the equation where it equals to 0.

$$9x^{3}+9x^{2}-x-1=0$$

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

$$±\frac{1}{9},±\frac{1}{3},±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 9x^{3}+9x^{2}-x-1 by x+1 to get 9x^{2}-1. To factor the result, solve the equation where it equals to 0.

$$9x^{2}-1=0$$

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 9 for a, 0 for b, and -1 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 9\left(-1\right)}}{2\times 9}$$

Do the calculations.

$$x=\frac{0±6}{18}$$

Solve the equation 9x^{2}-1=0 when ± is plus and when ± is minus.

$$x=-\frac{1}{3} x=\frac{1}{3}$$

Rewrite the factored expression using the obtained roots.

$$\left(x-1\right)\left(3x-1\right)\left(x+1\right)\left(3x+1\right)$$

Evaluate

$$9x^{4}-10x^{2}+1$$