Solve a^3+4+a^2=16

$$a ^ { 3 } + 4 + a ^ { 2 } = 16$$

Solve for a

$$a=2$$

Solution Steps

Subtract 16 from both sides.

$$a^{3}+4+a^{2}-16=0$$

Subtract 16 from 4 to get -12.

$$a^{3}-12+a^{2}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$a^{3}+a^{2}-12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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.

$$a=2$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{3}+a^{2}-12 by a-2 to get a^{2}+3a+6. Solve the equation where the result equals to 0.

$$a^{2}+3a+6=0$$

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 6 for c in the quadratic formula.

$$a=\frac{-3±\sqrt{3^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$a=\frac{-3±\sqrt{-15}}{2}$$

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

$$a\in \emptyset$$

List all found solutions.

$$a=2$$