Solve n^4-9n^2+4n^3-36n-12n^2+108

$$n ^ { 4 } - 9 n ^ { 2 } + 4 n ^ { 3 } - 36 n - 12 n ^ { 2 } + 108$$

Factor

$$\left(n-3\right)\left(n-2\right)\left(n+3\right)\left(n+6\right)$$

Solution Steps

Multiply and combine like terms.

$$n^{4}+4n^{3}-21n^{2}-36n+108$$

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

$$n^{4}+4n^{3}-21n^{2}-36n+108=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 108 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±108,±54,±36,±27,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$n=2$$

By Factor theorem, n-k is a factor of the polynomial for each root k. Divide n^{4}+4n^{3}-21n^{2}-36n+108 by n-2 to get n^{3}+6n^{2}-9n-54. To factor the result, solve the equation where it equals to 0.

$$n^{3}+6n^{2}-9n-54=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -54 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±54,±27,±18,±9,±6,±3,±2,±1$$

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.

$$n=3$$

By Factor theorem, n-k is a factor of the polynomial for each root k. Divide n^{3}+6n^{2}-9n-54 by n-3 to get n^{2}+9n+18. To factor the result, solve the equation where it equals to 0.

$$n^{2}+9n+18=0$$

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, 9 for b, and 18 for c in the quadratic formula.

$$n=\frac{-9±\sqrt{9^{2}-4\times 1\times 18}}{2}$$

Do the calculations.

$$n=\frac{-9±3}{2}$$

Solve the equation n^{2}+9n+18=0 when ± is plus and when ± is minus.

$$n=-6 n=-3$$

Rewrite the factored expression using the obtained roots.

$$\left(n-3\right)\left(n-2\right)\left(n+3\right)\left(n+6\right)$$

Evaluate

$$\left(n-2\right)\left(n+6\right)\left(n^{2}-9\right)$$

Solution Steps

Combine -9n^{2} and -12n^{2} to get -21n^{2}.

$$n^{4}-21n^{2}+4n^{3}-36n+108$$