Question

Divide using a long division method and check the answer.
\[{{x}^{2}}+5x+10\] by \[x+3\].

Answer 1

Hint: In this problem, we have to divide the given equation and the factor. We can use the polynomial long division method to divide the given problem by dividing the highest order term in the dividend to the highest order term in the divisor step by step until we get the remainder.

Complete step-by-step solution:
We know that the given division is,
\[\dfrac{{{x}^{2}}+5x+10}{x+3}\]
Now we can set up the polynomials to be divided in long division, we get
\[x+3\overset{{}}{\overline{\left){{{x}^{2}}+5x+10}\right.}}\]
Now we can divide the highest order term in the dividend \[{{x}^{2}}\] by the highest order term in the divisor x, we get
\[x+3\overset{x}{\overline{\left){{{x}^{2}}+5x+10}\right.}}\]
We can now multiply the quotient term to the divisor, we get
\[x+3\overset{x}{\overline{\left){\begin{align}
  & {{x}^{2}}+5x+10 \\
 & {{x}^{2}}+3x \\
\end{align}}\right.}}\]
We know that the expression is to be subtracted in the dividend, so we can change the sign in \[{{x}^{2}}+3x\], we get
\[\begin{align}
& x+3\underline{\overset{x}{\overline{\left){\begin{align}
& {{x}^{2}}+5x+10 \\
& -{{x}^{2}}-3x \\
\end{align}}\right.}}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x \\
\end{align}\]
Now we can bring down the next term from the dividend to the current dividend,
\[\begin{align}
& x+3\underline{\overset{x}{\overline{\left){\begin{align}
& {{x}^{2}}+5x+10 \\
& -{{x}^{2}}-3x \\
\end{align}}\right.}}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ 2}x+10 \\
\end{align}\]
Now we should divide the highest order term in the dividend by the divisor x, we get
\[\begin{align}
& x+3\underline{\overset{x+2}{\overline{\left){\begin{align}
& {{x}^{2}}+5x+10 \\
& -{{x}^{2}}-3x \\
\end{align}}\right.}}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ 2}x+10 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \text{ }\underline{2x+6} \\
\end{align}\]
We can now multiply the new quotient to the divisor, then subtract those expression to get the next dividend value, we get
\[\begin{align}
& x+3\underline{\overset{x+2}{\overline{\left){\begin{align}
& {{x}^{2}}+5x+10 \\
& -{{x}^{2}}-3x \\
\end{align}}\right.}}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{ 2}x+10 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{ }\underline{-2x-6} \\
\end{align}\]
Now we can subtract the last step by changing the signs, we get
\[\begin{align}
& x+3\underline{\overset{x+2}{\overline{\left){\begin{align}
& {{x}^{2}}+5x+10 \\
& -{{x}^{2}}-3x \\
\end{align}}\right.}}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ 2}x+10 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{ }\underline{-2x-6} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \text{ }4 \\
\end{align}\]
We know that dividend is equal to quotient plus the remainder over the divisor.
Therefore, the final answer is \[x+2+\dfrac{4}{x+3}\].

Note: We should also remember that the formula is Dividend = Quotient + Remainder/Divisor. Students make mistakes while finding the answer using polynomial long division. We should know how to find the answer using polynomial long division step by step. We will also make mistakes while changing the signs in order to cancel the step consequently.