Solve w^4-5w^3-5w^2-5w-6

$$w ^ { 4 } - 5 w ^ { 3 } - 5 w ^ { 2 } - 5 w - 6$$

Factor

$$\left(w-6\right)\left(w+1\right)\left(w^{2}+1\right)$$

Solution Steps

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

$$w^{4}-5w^{3}-5w^{2}-5w-6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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.

$$w=-1$$

By Factor theorem, w-k is a factor of the polynomial for each root k. Divide w^{4}-5w^{3}-5w^{2}-5w-6 by w+1 to get w^{3}-6w^{2}+w-6. To factor the result, solve the equation where it equals to 0.

$$w^{3}-6w^{2}+w-6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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.

$$w=6$$

By Factor theorem, w-k is a factor of the polynomial for each root k. Divide w^{3}-6w^{2}+w-6 by w-6 to get w^{2}+1. To factor the result, solve the equation where it equals to 0.

$$w^{2}+1=0$$

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 1 for a, 0 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$w=\frac{0±\sqrt{-4}}{2}$$

Polynomial w^{2}+1 is not factored since it does not have any rational roots.

$$w^{2}+1$$

Rewrite the factored expression using the obtained roots.

$$\left(w-6\right)\left(w+1\right)\left(w^{2}+1\right)$$

Evaluate

$$\left(w-6\right)\left(w+1\right)\left(w^{2}+1\right)$$