Question

Divide \(x^3 + 1\) by \(x + 1\).

Hint:

This problem can be solved instantly by using the sum of cubes factorization formula: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\). Here, \(x^3 + 1^3 = (x+1)(x^2 - x \cdot 1 + 1^2) = (x+1)(x^2-x+1)\). Therefore, \((x^3+1) \div (x+1) = x^2-x+1\).

Answer 1

Step 1: Understanding the Concept:
We can solve this problem using polynomial long division. It is important to write the dividend (\(x^3+1\)) with placeholders for the missing terms (\(x^2\) and \(x\)) to keep the columns aligned correctly.

Step 2: Detailed Explanation:
We set up the long division as follows, writing \(x^3 + 1\) as \(x^3 + 0x^2 + 0x + 1\).
\[ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & -x & +1
\cline{2-5} x+1 & x^3 & +0x^2 & +0x & +1
\multicolumn{2}{r}{- (x^3} & +x^2)
\cline{2-3} \multicolumn{2}{r}{0} & -x^2 & +0x
\multicolumn{2}{r}{} & -(-x^2 & -x)
\cline{3-4} \multicolumn{2}{r}{} & 0 & +x & +1
\multicolumn{2}{r}{} & & -(x & +1)
\cline{4-5} \multicolumn{2}{r}{} & & 0 & 0
\end{array} \] Step-by-step process:
\begin{enumerate} \item Divide the first term of the dividend (\(x^3\)) by the first term of the divisor (\(x\)). \(x^3/x = x^2\). Write \(x^2\) in the quotient. \item Multiply the divisor (\(x+1\)) by \(x^2\). \(x^2(x+1) = x^3+x^2\). Write this below the dividend. \item Subtract. \((x^3+0x^2) - (x^3+x^2) = -x^2\). Bring down the next term (\(0x\)). \item Divide the new first term (\(-x^2\)) by \(x\). \(-x^2/x = -x\). Write \(-x\) in the quotient. \item Multiply the divisor (\(x+1\)) by \(-x\). \(-x(x+1) = -x^2-x\). Write this below. \item Subtract. \((-x^2+0x) - (-x^2-x) = x\). Bring down the next term (\(+1\)). \item Divide the new first term (\(x\)) by \(x\). \(x/x = 1\). Write \(+1\) in the quotient. \item Multiply the divisor (\(x+1\)) by \(1\). \(1(x+1) = x+1\). Write this below. \item Subtract. \((x+1) - (x+1) = 0\). The remainder is 0. \end{enumerate}

Step 3: Final Answer:
The quotient is \(x^2 - x + 1\) and the remainder is 0.