Solve 4x^4-4x^3-51x^2+106x-40=0

$$4 { x }^{ 4 } -4 { x }^{ 3 } -51 { x }^{ 2 } +106x-40=0$$

Solve for x

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

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

$$x=2$$

$$x=-4$$

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 -40 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±10,±20,±40,±5,±\frac{5}{2},±2,±4,±8,±\frac{5}{4},±1,±\frac{1}{2},±\frac{1}{4}$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{4}-4x^{3}-51x^{2}+106x-40 by x-2 to get 4x^{3}+4x^{2}-43x+20. Solve the equation where the result equals to 0.

$$4x^{3}+4x^{2}-43x+20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 20 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±5,±10,±20,±\frac{5}{2},±\frac{5}{4},±1,±2,±4,±\frac{1}{2},±\frac{1}{4}$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+4x^{2}-43x+20 by x+4 to get 4x^{2}-12x+5. Solve the equation where the result equals to 0.

$$4x^{2}-12x+5=0$$

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, -12 for b, and 5 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 5}}{2\times 4}$$

Do the calculations.

$$x=\frac{12±8}{8}$$

Solve the equation 4x^{2}-12x+5=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=2 x=-4 x=\frac{1}{2} x=\frac{5}{2}$$