Solve 30x^3+3x^2+x=5

$$30 x ^ { 3 } + 3 x ^ { 2 } + x = 5$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{219}i}{30}-\frac{3}{10}\approx -0.3-0.493288286i$$

$$x=\frac{1}{2}=0.5$$

$$x=\frac{\sqrt{219}i}{30}-\frac{3}{10}\approx -0.3+0.493288286i$$

Solution Steps

Subtract 5 from both sides.

$$30x^{3}+3x^{2}+x-5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -5 and q divides the leading coefficient 30. List all candidates \frac{p}{q}.

$$±\frac{1}{6},±\frac{1}{3},±\frac{1}{2},±\frac{5}{6},±1,±\frac{5}{3},±\frac{5}{2},±5,±\frac{1}{30},±\frac{1}{15},±\frac{1}{10},±\frac{1}{5}$$

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.

$$x=\frac{1}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 30x^{3}+3x^{2}+x-5 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 15x^{2}+9x+5. Solve the equation where the result equals to 0.

$$15x^{2}+9x+5=0$$

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 15 for a, 9 for b, and 5 for c in the quadratic formula.

$$x=\frac{-9±\sqrt{9^{2}-4\times 15\times 5}}{2\times 15}$$

Do the calculations.

$$x=\frac{-9±\sqrt{-219}}{30}$$

Solve the equation 15x^{2}+9x+5=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{219}i}{30}-\frac{3}{10} x=\frac{\sqrt{219}i}{30}-\frac{3}{10}$$

List all found solutions.

$$x=\frac{1}{2} x=-\frac{\sqrt{219}i}{30}-\frac{3}{10} x=\frac{\sqrt{219}i}{30}-\frac{3}{10}$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Solution Steps

Subtract 5 from both sides.

$$30x^{3}+3x^{2}+x-5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -5 and q divides the leading coefficient 30. List all candidates \frac{p}{q}.

$$±\frac{1}{6},±\frac{1}{3},±\frac{1}{2},±\frac{5}{6},±1,±\frac{5}{3},±\frac{5}{2},±5,±\frac{1}{30},±\frac{1}{15},±\frac{1}{10},±\frac{1}{5}$$

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.

$$x=\frac{1}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 30x^{3}+3x^{2}+x-5 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 15x^{2}+9x+5. Solve the equation where the result equals to 0.

$$15x^{2}+9x+5=0$$

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 15 for a, 9 for b, and 5 for c in the quadratic formula.

$$x=\frac{-9±\sqrt{9^{2}-4\times 15\times 5}}{2\times 15}$$

Do the calculations.

$$x=\frac{-9±\sqrt{-219}}{30}$$

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.

$$x=\frac{1}{2}$$