Solve 50d^3+25d^2-2d-1=0

$$50 d ^ { 3 } + 25 d ^ { 2 } - 2 d - 1 = 0$$

Solve for d

$$d=-\frac{1}{2}=-0.5$$

$$d=\frac{1}{5}=0.2$$

$$d=-\frac{1}{5}=-0.2$$

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 -1 and q divides the leading coefficient 50. List all candidates \frac{p}{q}.

$$±\frac{1}{50},±\frac{1}{25},±\frac{1}{10},±\frac{1}{5},±\frac{1}{2},±1$$

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.

$$d=\frac{1}{5}$$

By Factor theorem, d-k is a factor of the polynomial for each root k. Divide 50d^{3}+25d^{2}-2d-1 by 5\left(d-\frac{1}{5}\right)=5d-1 to get 10d^{2}+7d+1. Solve the equation where the result equals to 0.

$$10d^{2}+7d+1=0$$

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 10 for a, 7 for b, and 1 for c in the quadratic formula.

$$d=\frac{-7±\sqrt{7^{2}-4\times 10\times 1}}{2\times 10}$$

Do the calculations.

$$d=\frac{-7±3}{20}$$

Solve the equation 10d^{2}+7d+1=0 when ± is plus and when ± is minus.

$$d=-\frac{1}{2} d=-\frac{1}{5}$$

List all found solutions.

$$d=\frac{1}{5} d=-\frac{1}{2} d=-\frac{1}{5}$$