What is the quotient of $\left(x^3+8\right) \div(x+2)$ ?
To find the quotient of \((x^3+8) \div (x+2)\), we first analyze the structure of the numerator. The expression \(x^3 + 8\) can be identified as a sum of cubes, because \(8 = 2^3\). Thus, the numerator is of the form \(x^3 + 2^3\).
We apply the sum of cubes factorization identity, which states that for any \(a\) and \(b\):
\[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]
In our case, let \(a = x\) and \(b = 2\). Substituting these values into the formula, we get:
\[x^3 + 2^3 = (x + 2)(x^2 - x(2) + 2^2)\]
Simplifying the expression inside the second parenthesis:
\[x^3 + 8 = (x + 2)(x^2 - 2x + 4)\]
Now we substitute the factored form of the numerator back into the original division problem:
\[\frac{x^3 + 8}{x + 2} = \frac{(x + 2)(x^2 - 2x + 4)}{x + 2}\]
Assuming \(x
eq -2\), we can cancel the common factor \((x + 2)\) from the numerator and the denominator:
\[\frac{(x + 2)(x^2 - 2x + 4)}{x + 2} = x^2 - 2x + 4\]
Thus, the quotient is \(x^2 - 2x + 4\).
Algebraic factorization is a fundamental tool used in many practical fields, such as engineering and physics, where complex equations need to be simplified to solve for specific variables. For instance, in structural analysis, engineers might simplify high-degree polynomial stress equations into lower-degree expressions to make them manageable. By factoring terms like sums of cubes, we can significantly reduce the complexity of mathematical models and gain insights into system behavior.