Solve 4x^3+x+4x^2-324=0

$$4 x ^ { 3 } + x + 4 x ^ { 2 } - 324 = 0$$

Solve for x (complex solution)

$$x=-\sqrt{14}i-\frac{5}{2}\approx -2.5-3.741657387i$$

$$x=4$$

$$x=-\frac{5}{2}+\sqrt{14}i\approx -2.5+3.741657387i$$

Solution Steps

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$4x^{3}+4x^{2}+x-324=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -324 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±81,±162,±324,±\frac{81}{2},±27,±54,±108,±\frac{81}{4},±\frac{27}{2},±9,±18,±36,±\frac{27}{4},±\frac{9}{2},±3,±6,±12,±\frac{9}{4},±\frac{3}{2},±1,±2,±4,±\frac{3}{4},±\frac{1}{2},±\frac{1}{4}$$

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 4x^{3}+4x^{2}+x-324 by x-4 to get 4x^{2}+20x+81. Solve the equation where the result equals to 0.

$$4x^{2}+20x+81=0$$

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 4 for a, 20 for b, and 81 for c in the quadratic formula.

$$x=\frac{-20±\sqrt{20^{2}-4\times 4\times 81}}{2\times 4}$$

Do the calculations.

$$x=\frac{-20±\sqrt{-896}}{8}$$

Solve the equation 4x^{2}+20x+81=0 when ± is plus and when ± is minus.

$$x=-\sqrt{14}i-\frac{5}{2} x=-\frac{5}{2}+\sqrt{14}i$$

List all found solutions.

$$x=4 x=-\sqrt{14}i-\frac{5}{2} x=-\frac{5}{2}+\sqrt{14}i$$

Solve for x

$$x=4$$

Solution Steps

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$4x^{3}+4x^{2}+x-324=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -324 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±81,±162,±324,±\frac{81}{2},±27,±54,±108,±\frac{81}{4},±\frac{27}{2},±9,±18,±36,±\frac{27}{4},±\frac{9}{2},±3,±6,±12,±\frac{9}{4},±\frac{3}{2},±1,±2,±4,±\frac{3}{4},±\frac{1}{2},±\frac{1}{4}$$

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 4x^{3}+4x^{2}+x-324 by x-4 to get 4x^{2}+20x+81. Solve the equation where the result equals to 0.

$$4x^{2}+20x+81=0$$

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 4 for a, 20 for b, and 81 for c in the quadratic formula.

$$x=\frac{-20±\sqrt{20^{2}-4\times 4\times 81}}{2\times 4}$$

Do the calculations.

$$x=\frac{-20±\sqrt{-896}}{8}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=4$$