Solve 4x^3+9x^2-10x-3=0

$$4 { x }^{ 3 } +9 { x }^{ 2 } -10x-3=0$$

Solve for x

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

$$x=1$$

$$x=-3$$

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 -3 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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}+9x^{2}-10x-3 by x-1 to get 4x^{2}+13x+3. Solve the equation where the result equals to 0.

$$4x^{2}+13x+3=0$$

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, 13 for b, and 3 for c in the quadratic formula.

$$x=\frac{-13±\sqrt{13^{2}-4\times 4\times 3}}{2\times 4}$$

Do the calculations.

$$x=\frac{-13±11}{8}$$

Solve the equation 4x^{2}+13x+3=0 when ± is plus and when ± is minus.

$$x=-3 x=-\frac{1}{4}$$

List all found solutions.

$$x=1 x=-3 x=-\frac{1}{4}$$