Solve x^3-3x^2-49x+147=0

$$x ^ { 3 } - 3 x ^ { 2 } - 49 x + 147 = 0$$

Solve for x

$$x=3$$

$$x=7$$

$$x=-7$$

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 147 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±147,±49,±21,±7,±3,±1$$

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 x^{3}-3x^{2}-49x+147 by x-3 to get x^{2}-49. Solve the equation where the result equals to 0.

$$x^{2}-49=0$$

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 1 for a, 0 for b, and -49 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-49\right)}}{2}$$

Do the calculations.

$$x=\frac{0±14}{2}$$

Solve the equation x^{2}-49=0 when ± is plus and when ± is minus.

$$x=-7 x=7$$

List all found solutions.

$$x=3 x=-7 x=7$$