Solve 4x^3+8x^2+5x+3=0

$$4 x ^ { 3 } + 8 x ^ { 2 } + 5 x + 3 = 0$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{7}i}{4}\approx -0.25+0.661437828i$$

$$x=\frac{-\sqrt{7}i-1}{4}\approx -0.25-0.661437828i$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Solution Steps

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

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

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.

$$x=-\frac{3}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+8x^{2}+5x+3 by 2\left(x+\frac{3}{2}\right)=2x+3 to get 2x^{2}+x+1. Solve the equation where the result equals to 0.

$$2x^{2}+x+1=0$$

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 2 for a, 1 for b, and 1 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-7}}{4}$$

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

$$x=\frac{-\sqrt{7}i-1}{4} x=\frac{-1+\sqrt{7}i}{4}$$

List all found solutions.

$$x=-\frac{3}{2} x=\frac{-\sqrt{7}i-1}{4} x=\frac{-1+\sqrt{7}i}{4}$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Solution Steps

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

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

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.

$$x=-\frac{3}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+8x^{2}+5x+3 by 2\left(x+\frac{3}{2}\right)=2x+3 to get 2x^{2}+x+1. Solve the equation where the result equals to 0.

$$2x^{2}+x+1=0$$

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 2 for a, 1 for b, and 1 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-7}}{4}$$

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.

$$x=-\frac{3}{2}$$