Example

Divide: [latex]\frac{\left(x^{2}–4x–12\right)}{\left(x+2\right)}[/latex]

Answer: How many x’s are there in [latex]x^{2}[/latex]? That is, what is [latex] \frac{{{x}^{2}}}{x}[/latex]? [latex] \frac{{{x}^{2}}}{x}=x[/latex]. Put x in the quotient above the [latex]-4x[/latex] term. These are like terms, which helps to organize the problem. Write the product of the divisor and the part of the quotient you just found under the dividend. Since [latex]x\left(x+2\right)=x^{2}+2x[/latex], write this underneath, and get ready to subtract. Rewrite [latex]–\left(x^{2} + 2x\right)[/latex] as its opposite [latex]–x^{2}–2x[/latex] so that you can add the opposite. Adding the opposite is the same as subtracting, and it is easier to do. Add [latex]-x^{2}[/latex] to [latex]x^{2}[/latex] and [latex]-2x[/latex] to [latex]-4x[/latex]. Bring down [latex]-12[/latex]. Repeat the process. How many times does x go into [latex]-6x[/latex]? In other words, what is [latex] \frac{-6x}{x}[/latex]? Since [latex] \frac{-6x}{x}=-6[/latex], write [latex]-6[/latex] in the quotient. Multiply [latex]-6[/latex] and [latex]x+2[/latex] and prepare to subtract the product. Rewrite [latex]–\left(-6x–12\right)[/latex] as [latex]6x+12[/latex], so that you can add the opposite. Add. In this case, there is no remainder, so you are done. The answer is [latex]x–6[/latex]. Check this by multiplying:

[latex]\left(x-6\right)\left(x+2\right)=x^{2}+2x-6x-12=x^{2}-4x-12[/latex]

Example

Divide: [latex]\frac{\left(x^{3}–6x–10\right)}{\left(x–3\right)}[/latex]

Answer: In setting up this problem, notice that there is an [latex]x^{3}[/latex] term but no [latex]x^{2}[/latex] term. Add [latex]0x^{2}[/latex] as a “place holder” for this term. Since 0 times anything is 0, you are not changing the value of the dividend. Focus on the first terms again: how many x’s are there in [latex]x^{3}[/latex]? Since [latex] \frac{{{x}^{3}}}{x}=x^{2}[/latex], put [latex]x^{2}[/latex] in the quotient. Multiply: [latex]x^{2}\left(x–3\right)=x^{3}–3x^{2}[/latex]; write this underneath the dividend, and prepare to subtract. Rewrite the subtraction using the opposite of the expression [latex]x^{3}-3x^{2}[/latex]. Then add. Bring down the rest of the expression in the dividend. It is helpful to bring down all of the remaining terms. Now, repeat the process with the remaining expression, [latex]3x^{2}-6x–10[/latex], as the dividend. Remember to watch the signs! How many x’s are there in [latex]3[/latex]x? Since there are [latex]3[/latex], multiply [latex]3\left(x–3\right)=3x–9[/latex], write this underneath the dividend, and prepare to subtract. Continue until the degree of the remainder is less than the degree of the divisor. In this case the degree of the remainder, [latex]-1[/latex], is [latex]0[/latex], which is less than the degree of [latex]x-3[/latex], which is [latex]1[/latex]. Also notice that you have brought down all the terms in the dividend and that the quotient extends to the right edge of the dividend. These are other ways to check whether you have completed the problem. You can write the remainder using the symbol R or as a fraction added to the rest of the quotient with the remainder in the numerator and the divisor in the denominator. In this case, since the remainder is negative, you can also subtract the opposite. Possible forms of the solution are: [latex-display]\begin{array}{l}x^{2}+3x+3+R-1,\\x^{2}+3x+3+\frac{-1}{x-3}, \text{ or }\\x^{2}+3x+3-\frac{1}{x-3}\end{array}[/latex-display] Check the result: [latex-display]\begin{array}{l}\left(x–3\right)\left(x^{2}+3x+3\right) & =x\left(x^{2}+3x+3\right)–3\left(x^{2}+3x+3\right)\\ & =x^{3}+3x^{2}+3x–3x^{2}–9x–9\\ & =x^{3}–6x–9\end{array}[/latex-display] [latex-display]x^{3}–6x–9+\left(-1\right)=x^{3}–6x–10[/latex-display]