Solve 8x^3-36x^2+46x-15=0

$$8 x ^ { 3 } - 36 x ^ { 2 } + 46 x - 15 = 0$$

Solve for x

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

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

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

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 -15 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{15}{8},±\frac{15}{4},±\frac{15}{2},±15,±\frac{5}{8},±\frac{5}{4},±\frac{5}{2},±5,±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}-36x^{2}+46x-15 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 4x^{2}-16x+15. Solve the equation where the result equals to 0.

$$4x^{2}-16x+15=0$$

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 4 for a, -16 for b, and 15 for c in the quadratic formula.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\times 15}}{2\times 4}$$

Do the calculations.

$$x=\frac{16±4}{8}$$

Solve the equation 4x^{2}-16x+15=0 when ± is plus and when ± is minus.

$$x=\frac{3}{2} x=\frac{5}{2}$$

List all found solutions.

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