Solve 8x^3+15x^2-36x-68=0

$$8 x ^ { 3 } + 15 x ^ { 2 } - 36 x - 68 = 0$$

Solve for x

$$x=-2$$

$$x = \frac{17}{8} = 2\frac{1}{8} = 2.125$$

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 -68 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{17}{2},±17,±34,±68,±\frac{17}{4},±\frac{17}{8},±\frac{1}{2},±1,±2,±4,±\frac{1}{4},±\frac{1}{8}$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}+15x^{2}-36x-68 by x+2 to get 8x^{2}-x-34. Solve the equation where the result equals to 0.

$$8x^{2}-x-34=0$$

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 8 for a, -1 for b, and -34 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 8\left(-34\right)}}{2\times 8}$$

Do the calculations.

$$x=\frac{1±33}{16}$$

Solve the equation 8x^{2}-x-34=0 when ± is plus and when ± is minus.

$$x=-2 x=\frac{17}{8}$$

List all found solutions.

$$x=-2 x=\frac{17}{8}$$