Solve 3x^3+x^2-19x+15=0

$$3 x ^ { 3 } + x ^ { 2 } - 19 x + 15 = 0$$

Solve for x

$$x=1$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=-3$$

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 15 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±5,±15,±\frac{5}{3},±1,±3,±\frac{1}{3}$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}+x^{2}-19x+15 by x-1 to get 3x^{2}+4x-15. Solve the equation where the result equals to 0.

$$3x^{2}+4x-15=0$$

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, 4 for b, and -15 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{-4±14}{6}$$

Solve the equation 3x^{2}+4x-15=0 when ± is plus and when ± is minus.

$$x=-3 x=\frac{5}{3}$$

List all found solutions.

$$x=1 x=-3 x=\frac{5}{3}$$