Solve =20x^4+31x^2-9

$$= 20 x ^ { 4 } + 31 x ^ { 2 } - 9$$

Factor

$$\left(2x-1\right)\left(2x+1\right)\left(5x^{2}+9\right)$$

Solution Steps

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

$$20x^{4}+31x^{2}-9=0$$

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

$$±\frac{9}{20},±\frac{9}{10},±\frac{9}{5},±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{20},±\frac{3}{10},±\frac{3}{5},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{20},±\frac{1}{10},±\frac{1}{5},±\frac{1}{4},±\frac{1}{2},±1$$

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 20x^{4}+31x^{2}-9 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 10x^{3}+5x^{2}+18x+9. To factor the result, solve the equation where it equals to 0.

$$10x^{3}+5x^{2}+18x+9=0$$

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

$$±\frac{9}{10},±\frac{9}{5},±\frac{9}{2},±9,±\frac{3}{10},±\frac{3}{5},±\frac{3}{2},±3,±\frac{1}{10},±\frac{1}{5},±\frac{1}{2},±1$$

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 10x^{3}+5x^{2}+18x+9 by 2\left(x+\frac{1}{2}\right)=2x+1 to get 5x^{2}+9. To factor the result, solve the equation where it equals to 0.

$$5x^{2}+9=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\times 5\times 9}}{2\times 5}$$

Do the calculations.

$$x=\frac{0±\sqrt{-180}}{10}$$

Polynomial 5x^{2}+9 is not factored since it does not have any rational roots.

$$5x^{2}+9$$

Rewrite the factored expression using the obtained roots.

$$\left(2x-1\right)\left(2x+1\right)\left(5x^{2}+9\right)$$

Evaluate

$$20x^{4}+31x^{2}-9$$