Solve 3x^4+24x^3+48x^2-x^2-8x-16

$$3 x ^ { 4 } + 24 x ^ { 3 } + 48 x ^ { 2 } - x ^ { 2 } - 8 x - 16$$

Factor

$$\left(3x^{2}-1\right)\left(x+4\right)^{2}$$

Solution Steps

Multiply and combine like terms.

$$3x^{4}+24x^{3}+47x^{2}-8x-16$$

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

$$3x^{4}+24x^{3}+47x^{2}-8x-16=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -16 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{16}{3},±16,±\frac{8}{3},±8,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{4}+24x^{3}+47x^{2}-8x-16 by x+4 to get 3x^{3}+12x^{2}-x-4. To factor the result, solve the equation where it equals to 0.

$$3x^{3}+12x^{2}-x-4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}+12x^{2}-x-4 by x+4 to get 3x^{2}-1. To factor the result, solve the equation where it equals to 0.

$$3x^{2}-1=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{0±2\sqrt{3}}{6}$$

Solve the equation 3x^{2}-1=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{3}}{3} x=\frac{\sqrt{3}}{3}$$

Rewrite the factored expression using the obtained roots. Polynomial 3x^{2}-1 is not factored since it does not have any rational roots.

$$\left(3x^{2}-1\right)\left(x+4\right)^{2}$$

Evaluate

$$\left(3x^{2}-1\right)\left(x+4\right)^{2}$$

Solution Steps

Combine 48x^{2} and -x^{2} to get 47x^{2}.

$$3x^{4}+24x^{3}+47x^{2}-8x-16$$