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

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

Solve for x

$$x=-1$$

$$x=1$$

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

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

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

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 4x^{3}+5x^{2}-4x-5 by x-1 to get 4x^{2}+9x+5. Solve the equation where the result equals to 0.

$$4x^{2}+9x+5=0$$

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, 9 for b, and 5 for c in the quadratic formula.

$$x=\frac{-9±\sqrt{9^{2}-4\times 4\times 5}}{2\times 4}$$

Do the calculations.

$$x=\frac{-9±1}{8}$$

Solve the equation 4x^{2}+9x+5=0 when ± is plus and when ± is minus.

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

List all found solutions.

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