Solve 12x^3-28x^2-3x+7=0

$$12 x ^ { 3 } - 28 x ^ { 2 } - 3 x + 7 = 0$$

Solve for x

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

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

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

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 7 and q divides the leading coefficient 12. List all candidates \frac{p}{q}.

$$±\frac{7}{12},±\frac{7}{6},±\frac{7}{4},±\frac{7}{3},±\frac{7}{2},±7,±\frac{1}{12},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 12x^{3}-28x^{2}-3x+7 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 6x^{2}-11x-7. Solve the equation where the result equals to 0.

$$6x^{2}-11x-7=0$$

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 6 for a, -11 for b, and -7 for c in the quadratic formula.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\left(-7\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{11±17}{12}$$

Solve the equation 6x^{2}-11x-7=0 when ± is plus and when ± is minus.

$$x=-\frac{1}{2} x=\frac{7}{3}$$

List all found solutions.

$$x=\frac{1}{2} x=-\frac{1}{2} x=\frac{7}{3}$$