Solve 12x^3+4x^2-7x+1=0

$$12 x ^ { 3 } + 4 x ^ { 2 } - 7 x + 1 = 0$$

Solve for x

$$x=\frac{1}{6}\approx 0.166666667$$

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

$$x=-1$$

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 12. List all candidates \frac{p}{q}.

$$±\frac{1}{12},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 12x^{3}+4x^{2}-7x+1 by x+1 to get 12x^{2}-8x+1. Solve the equation where the result equals to 0.

$$12x^{2}-8x+1=0$$

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

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 12\times 1}}{2\times 12}$$

Do the calculations.

$$x=\frac{8±4}{24}$$

Solve the equation 12x^{2}-8x+1=0 when ± is plus and when ± is minus.

$$x=\frac{1}{6} x=\frac{1}{2}$$

List all found solutions.

$$x=-1 x=\frac{1}{6} x=\frac{1}{2}$$