Solve f(x)=x^5-7x^4-13x^3-37x^2-68x-36

$$f ( x ) = x ^ { 5 } - 7 x ^ { 4 } - 13 x ^ { 3 } - 37 x ^ { 2 } - 68 x - 36$$

Factor

$$\left(x-9\right)\left(x+1\right)^{2}\left(x^{2}+4\right)$$

Solution Steps

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

$$x^{5}-7x^{4}-13x^{3}-37x^{2}-68x-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{5}-7x^{4}-13x^{3}-37x^{2}-68x-36 by x+1 to get x^{4}-8x^{3}-5x^{2}-32x-36. To factor the result, solve the equation where it equals to 0.

$$x^{4}-8x^{3}-5x^{2}-32x-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-8x^{3}-5x^{2}-32x-36 by x+1 to get x^{3}-9x^{2}+4x-36. To factor the result, solve the equation where it equals to 0.

$$x^{3}-9x^{2}+4x-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$x=9$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+4x-36 by x-9 to get x^{2}+4. To factor the result, solve the equation where it equals to 0.

$$x^{2}+4=0$$

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 4 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-16}}{2}$$

Polynomial x^{2}+4 is not factored since it does not have any rational roots.

$$x^{2}+4$$

Rewrite the factored expression using the obtained roots.

$$\left(x-9\right)\left(x+1\right)^{2}\left(x^{2}+4\right)$$

Evaluate

$$\left(x-9\right)\left(x+1\right)^{2}\left(x^{2}+4\right)$$