Solve b^4-10b^2+9

$$b ^ { 4 } - 10 b ^ { 2 } + 9$$

Factor

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

Solution Steps

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

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

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

$$±9,±3,±1$$

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.

$$b=1$$

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

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

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

$$±9,±3,±1$$

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.

$$b=-1$$

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

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

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 -9 for c in the quadratic formula.

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

Do the calculations.

$$b=\frac{0±6}{2}$$

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

$$b=-3 b=3$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

$$b^{4}-10b^{2}+9$$