Solve f(x)=5x^4+2x^2-7

$$f ( x ) = 5 x ^ { 4 } + 2 x ^ { 2 } - 7$$

Factor

$$\left(x-1\right)\left(x+1\right)\left(5x^{2}+7\right)$$

Solution Steps

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

$$5x^{4}+2x^{2}-7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -7 and q divides the leading coefficient 5. List all candidates \frac{p}{q}.

$$±\frac{7}{5},±7,±\frac{1}{5},±1$$

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 5x^{4}+2x^{2}-7 by x-1 to get 5x^{3}+5x^{2}+7x+7. To factor the result, solve the equation where it equals to 0.

$$5x^{3}+5x^{2}+7x+7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 7 and q divides the leading coefficient 5. List all candidates \frac{p}{q}.

$$±\frac{7}{5},±7,±\frac{1}{5},±1$$

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 5x^{3}+5x^{2}+7x+7 by x+1 to get 5x^{2}+7. To factor the result, solve the equation where it equals to 0.

$$5x^{2}+7=0$$

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 7 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 5\times 7}}{2\times 5}$$

Do the calculations.

$$x=\frac{0±\sqrt{-140}}{10}$$

Polynomial 5x^{2}+7 is not factored since it does not have any rational roots.

$$5x^{2}+7$$

Rewrite the factored expression using the obtained roots.

$$\left(x-1\right)\left(x+1\right)\left(5x^{2}+7\right)$$

Evaluate

$$5x^{4}+2x^{2}-7$$