Solve 5x^3-4x^2-5x+4=0

$$5 x ^ { 3 } - 4 x ^ { 2 } - 5 x + 4 = 0$$

Solve for x

$$x=1$$

$$x=\frac{4}{5}=0.8$$

$$x=-1$$

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 4 and q divides the leading coefficient 5. List all candidates \frac{p}{q}.

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

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 5x^{3}-4x^{2}-5x+4 by x-1 to get 5x^{2}+x-4. Solve the equation where the result equals to 0.

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

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 5 for a, 1 for b, and -4 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 5\left(-4\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{-1±9}{10}$$

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

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

List all found solutions.

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