Solve 4x^3+7x^2-36x-63=0

$$4 x ^ { 3 } + 7 x ^ { 2 } - 36 x - 63 = 0$$

Solve for x

$$x=-3$$

$$x=3$$

$$x = -\frac{7}{4} = -1\frac{3}{4} = -1.75$$

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 -63 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{63}{4},±\frac{63}{2},±63,±\frac{21}{4},±\frac{21}{2},±21,±\frac{9}{4},±\frac{9}{2},±9,±\frac{7}{4},±\frac{7}{2},±7,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+7x^{2}-36x-63 by x-3 to get 4x^{2}+19x+21. Solve the equation where the result equals to 0.

$$4x^{2}+19x+21=0$$

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, 19 for b, and 21 for c in the quadratic formula.

$$x=\frac{-19±\sqrt{19^{2}-4\times 4\times 21}}{2\times 4}$$

Do the calculations.

$$x=\frac{-19±5}{8}$$

Solve the equation 4x^{2}+19x+21=0 when ± is plus and when ± is minus.

$$x=-3 x=-\frac{7}{4}$$

List all found solutions.

$$x=3 x=-3 x=-\frac{7}{4}$$