Solve 4x^3-21x^2+21x-4=0

$$4 { x }^{ 3 } -21 { x }^{ 2 } +21x-4=0$$

Solve for x

$$x=1$$

$$x=4$$

$$x=\frac{1}{4}=0.25$$

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 -4 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±1,±2,±4,±\frac{1}{2},±\frac{1}{4}$$

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 4x^{3}-21x^{2}+21x-4 by x-1 to get 4x^{2}-17x+4. Solve the equation where the result equals to 0.

$$4x^{2}-17x+4=0$$

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, -17 for b, and 4 for c in the quadratic formula.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Do the calculations.

$$x=\frac{17±15}{8}$$

Solve the equation 4x^{2}-17x+4=0 when ± is plus and when ± is minus.

$$x=\frac{1}{4} x=4$$

List all found solutions.

$$x=1 x=\frac{1}{4} x=4$$