Solve 2p_2^3=54

$$2 p _ { 2 } ^ { 3 } = 54$$

Solve for p_2

$$p_{2}=3$$

Solution Steps

Divide both sides by 2.

$$p_{2}^{3}=\frac{54}{2}$$

Divide 54 by 2 to get 27.

$$p_{2}^{3}=27$$

Subtract 27 from both sides.

$$p_{2}^{3}-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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.

$$p_{2}=3$$

By Factor theorem, p_{2}-k is a factor of the polynomial for each root k. Divide p_{2}^{3}-27 by p_{2}-3 to get p_{2}^{2}+3p_{2}+9. Solve the equation where the result equals to 0.

$$p_{2}^{2}+3p_{2}+9=0$$

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

$$p_{2}=\frac{-3±\sqrt{3^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$p_{2}=\frac{-3±\sqrt{-27}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$p_{2}\in \emptyset$$

List all found solutions.

$$p_{2}=3$$