
How do you long divide $\dfrac{{{x}^{2}}+x-17}{x-4}$ ?
Answer
497.4k+ views
Hint: In the above problem we are asked to long divide ${{x}^{2}}+x-17$ by $x-4$. For that, we will first of all multiply by such an expression to $x-4$ so that at least ${{x}^{2}}$ will vanish. Then we are going to write the result of multiplication of $x-4$ by some expression below ${{x}^{2}}+x$ and then subtract them. After doing the subtraction, we are going to carry -17 from ${{x}^{2}}+x-17$ and will write just after the subtraction. And then again, we will multiply $x-4$ by such an expression so that the remaining expression will vanish. From this, you will get quotient and remainder.
Complete step by step answer:
In the below, we are going to long divide ${{x}^{2}}+x-17$ by $x-4$ as follows:
$x-4\overset{{}}{\overline{\left){{{x}^{2}}+x-17}\right.}}$
Now, we are going to multiply $x-4$ by $x$ so that ${{x}^{2}}$ will vanish by subtracting the result of the multiplication of $x-4$ by $x$.
$x-4\overset{x}{\overline{\left){\begin{align}
& {{x}^{2}}+x-17 \\
& \dfrac{\begin{align}
& {{x}^{2}}-4x \\
& \begin{matrix}
- & + \\
\end{matrix} \\
\end{align}}{0+5x} \\
\end{align}}\right.}}$
Moving further in the above long division we are going to carry -17 from ${{x}^{2}}+x-17$ and will write -17 just after $5x$.
\[x-4\overset{x}{\overline{\left){\begin{align}
& {{x}^{2}}+x-17 \\
& {{x}^{2}}-4x \\
& \begin{matrix}
- & + \\
\end{matrix} \\
& \overline{0+5x-17} \\
\end{align}}\right.}}\]
Now, we are going to multiply $x-4$ by 5 we get,
\[x-4\overset{x+5}{\overline{\left){\begin{align}
& {{x}^{2}}+x-17 \\
& {{x}^{2}}-4x \\
& \begin{matrix}
- & + \\
\end{matrix} \\
& \overline{\begin{align}
& 0+5x-17 \\
& 0+5x-20 \\
\end{align}} \\
\end{align}}\right.}}\]
Subtracting $5x-20$ from $5x-17$ we get,
\[x-4\overset{x+5}{\overline{\left){\begin{align}
& {{x}^{2}}+x-17 \\
& {{x}^{2}}-4x \\
& \begin{matrix}
- & + \\
\end{matrix} \\
& \overline{\begin{align}
& 0+5x-17 \\
& \dfrac{\begin{align}
& 0+5x-20 \\
& \begin{matrix}
- & - & + \\
\end{matrix} \\
\end{align}}{0+0+3} \\
\end{align}} \\
\end{align}}\right.}}\]
From the above long division, we got the quotient as $x+5$ and remainder as 3.
Hence, we have long divided ${{x}^{2}}+x-17$ by $x-4$ and we got the quotient as $x+5$ and remainder as 3.
Note: We can check the quotient and remainder that we got from the long division is correct or not by using the following relation:
\[\text{Dividend}=\text{Divisor}\times \text{Quotient}+\text{Remainder}\]
Now, dividend is ${{x}^{2}}+x-17$, divisor is $x-4$, quotient is $x+5$ and remainder is 3 so substituting these values in the above relation we get,
\[{{x}^{2}}+x-17=\left( x-4 \right)\times \left( x+5 \right)+3\]
Multiplying $\left( x-4 \right)$ by $\left( x+5 \right)$ in the above equation we get,
\[\begin{align}
& {{x}^{2}}+x-17=\left( {{x}^{2}}+5x-4x-20 \right)+3 \\
& \Rightarrow {{x}^{2}}+x-17={{x}^{2}}+x-20+3 \\
& \Rightarrow {{x}^{2}}+x-17={{x}^{2}}+x-17 \\
\end{align}\]
As you can see that L.H.S is equal to R.H.S so the long division that we did is correct.
Complete step by step answer:
In the below, we are going to long divide ${{x}^{2}}+x-17$ by $x-4$ as follows:
$x-4\overset{{}}{\overline{\left){{{x}^{2}}+x-17}\right.}}$
Now, we are going to multiply $x-4$ by $x$ so that ${{x}^{2}}$ will vanish by subtracting the result of the multiplication of $x-4$ by $x$.
$x-4\overset{x}{\overline{\left){\begin{align}
& {{x}^{2}}+x-17 \\
& \dfrac{\begin{align}
& {{x}^{2}}-4x \\
& \begin{matrix}
- & + \\
\end{matrix} \\
\end{align}}{0+5x} \\
\end{align}}\right.}}$
Moving further in the above long division we are going to carry -17 from ${{x}^{2}}+x-17$ and will write -17 just after $5x$.
\[x-4\overset{x}{\overline{\left){\begin{align}
& {{x}^{2}}+x-17 \\
& {{x}^{2}}-4x \\
& \begin{matrix}
- & + \\
\end{matrix} \\
& \overline{0+5x-17} \\
\end{align}}\right.}}\]
Now, we are going to multiply $x-4$ by 5 we get,
\[x-4\overset{x+5}{\overline{\left){\begin{align}
& {{x}^{2}}+x-17 \\
& {{x}^{2}}-4x \\
& \begin{matrix}
- & + \\
\end{matrix} \\
& \overline{\begin{align}
& 0+5x-17 \\
& 0+5x-20 \\
\end{align}} \\
\end{align}}\right.}}\]
Subtracting $5x-20$ from $5x-17$ we get,
\[x-4\overset{x+5}{\overline{\left){\begin{align}
& {{x}^{2}}+x-17 \\
& {{x}^{2}}-4x \\
& \begin{matrix}
- & + \\
\end{matrix} \\
& \overline{\begin{align}
& 0+5x-17 \\
& \dfrac{\begin{align}
& 0+5x-20 \\
& \begin{matrix}
- & - & + \\
\end{matrix} \\
\end{align}}{0+0+3} \\
\end{align}} \\
\end{align}}\right.}}\]
From the above long division, we got the quotient as $x+5$ and remainder as 3.
Hence, we have long divided ${{x}^{2}}+x-17$ by $x-4$ and we got the quotient as $x+5$ and remainder as 3.
Note: We can check the quotient and remainder that we got from the long division is correct or not by using the following relation:
\[\text{Dividend}=\text{Divisor}\times \text{Quotient}+\text{Remainder}\]
Now, dividend is ${{x}^{2}}+x-17$, divisor is $x-4$, quotient is $x+5$ and remainder is 3 so substituting these values in the above relation we get,
\[{{x}^{2}}+x-17=\left( x-4 \right)\times \left( x+5 \right)+3\]
Multiplying $\left( x-4 \right)$ by $\left( x+5 \right)$ in the above equation we get,
\[\begin{align}
& {{x}^{2}}+x-17=\left( {{x}^{2}}+5x-4x-20 \right)+3 \\
& \Rightarrow {{x}^{2}}+x-17={{x}^{2}}+x-20+3 \\
& \Rightarrow {{x}^{2}}+x-17={{x}^{2}}+x-17 \\
\end{align}\]
As you can see that L.H.S is equal to R.H.S so the long division that we did is correct.
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