Solve 9b^4-82b^2+9

$$9 b ^ { 4 } - 82 b ^ { 2 } + 9$$

Factor

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

Solution Steps

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

$$9b^{4}-82b^{2}+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 9. List all candidates \frac{p}{q}.

$$±1,±3,±9,±\frac{1}{3},±\frac{1}{9}$$

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.

$$b=3$$

By Factor theorem, b-k is a factor of the polynomial for each root k. Divide 9b^{4}-82b^{2}+9 by b-3 to get 9b^{3}+27b^{2}-b-3. To factor the result, solve the equation where it equals to 0.

$$9b^{3}+27b^{2}-b-3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -3 and q divides the leading coefficient 9. List all candidates \frac{p}{q}.

$$±\frac{1}{3},±1,±3,±\frac{1}{9}$$

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.

$$b=-3$$

By Factor theorem, b-k is a factor of the polynomial for each root k. Divide 9b^{3}+27b^{2}-b-3 by b+3 to get 9b^{2}-1. To factor the result, solve the equation where it equals to 0.

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

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.

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

Do the calculations.

$$b=\frac{0±6}{18}$$

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

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

Rewrite the factored expression using the obtained roots.

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

Evaluate

$$9b^{4}-82b^{2}+9$$