Solve =x^4-x^3-7x^2+x+6

$$= x ^ { 4 } - x ^ { 3 } - 7 x ^ { 2 } + x + 6$$

Factor

$$\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)$$

Solution Steps

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

$$x^{4}-x^{3}-7x^{2}+x+6=0$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-x^{3}-7x^{2}+x+6 by x-1 to get x^{3}-7x-6. To factor the result, solve the equation where it equals to 0.

$$x^{3}-7x-6=0$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-7x-6 by x+1 to get x^{2}-x-6. To factor the result, solve the equation where it equals to 0.

$$x^{2}-x-6=0$$

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, -1 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-6\right)}}{2}$$

Do the calculations.

$$x=\frac{1±5}{2}$$

Solve the equation x^{2}-x-6=0 when ± is plus and when ± is minus.

$$x=-2 x=3$$

Rewrite the factored expression using the obtained roots.

$$\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)$$

Evaluate

$$\left(x-3\right)\left(x+2\right)\left(x^{2}-1\right)$$