
How do you divide \[\left( {{x}^{4}}-16 \right)\div \left( x+2 \right)\] using synthetic division?
Answer
520.5k+ views
Hint: In this problem, we have to divide the given equation and the factor using the synthetic division. By using a synthetic division method, we can place the numbers from the given equation in a L-shaped configuration. We can solve them step by step to get the quotient and the remainder.
Complete step-by-step solution:
We know that the given division to be divided is,
\[\left( {{x}^{4}}-16 \right)\div \left( x+2 \right)\].
We can see that the given factor as a divisor is \[\left( x+2 \right)\].
We can write it as x = -2, which is the divisor in the synthetic division.
We can now place the numbers from the given equation in a L-shaped configuration.
\[-2\underline{\left| 1 \right.\text{ 0 0 0 }-16}\]
Here we have added zeros, as we don’t have those terms.
We can put the first number in the dividend in the first position of the result area, we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\; 1 \\
\end{align}\]
Now we can multiply the newest entry (1) by the divisor (-2) and place the result of (-2) under the next term in the dividend (0), we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\; 1 \\
\end{align}\]
Now we can add 0 and -2 and put the result in the next position, we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\; 1 \;\;\; -2 \\
\end{align}\]
Now we can multiply the newest entry (-2) by the divisor (-2) and place the result of (4) under the next term in the dividend (0), we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2\text{ }4 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\;1\;\;\; -2 \\
\end{align}\]
Now we can add 0 and 4 and put the result in the next position, we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2\text{ }4 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\;1\;\;\; -2\;\;\;4 \\
\end{align}\]
Now we can multiply the newest entry (4) by the divisor (-2) and place the result of (-8) under the next term in the dividend (0), we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2\text{ }4\text{ }-8 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\;1\;\;\;\; -2\;\;\;\ 4\;\;\;\ -8 \\
\end{align}\]
Now we can multiply the newest entry (-8) by the divisor (-2) and place the result of (16) under the next term in the dividend (-16), we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2\text{ }4\text{ }-8\text{ }16 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\;1\;\;\;\; -2\;\;\; 4\;\;\;\; -8\;\;\; 0 \\
\end{align}\]
We know that dividend is equal to quotient plus the remainder over the divisor.
Dividend = Quotient + Remainder/Divisor.
Therefore, the final answer is \[{{x}^{3}}-2{{x}^{2}}+4x-8\].
Note: Students make mistakes while using the synthetic division. We should first differentiate the dividend and the divisor. We should take the first term of the dividend to the result and we should start multiplying it with the divisor to put it for the next value which is added to the next term of the divisor. By repeating this method, we will get the quotient and the remainder value.
Complete step-by-step solution:
We know that the given division to be divided is,
\[\left( {{x}^{4}}-16 \right)\div \left( x+2 \right)\].
We can see that the given factor as a divisor is \[\left( x+2 \right)\].
We can write it as x = -2, which is the divisor in the synthetic division.
We can now place the numbers from the given equation in a L-shaped configuration.
\[-2\underline{\left| 1 \right.\text{ 0 0 0 }-16}\]
Here we have added zeros, as we don’t have those terms.
We can put the first number in the dividend in the first position of the result area, we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\; 1 \\
\end{align}\]
Now we can multiply the newest entry (1) by the divisor (-2) and place the result of (-2) under the next term in the dividend (0), we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\; 1 \\
\end{align}\]
Now we can add 0 and -2 and put the result in the next position, we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\; 1 \;\;\; -2 \\
\end{align}\]
Now we can multiply the newest entry (-2) by the divisor (-2) and place the result of (4) under the next term in the dividend (0), we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2\text{ }4 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\;1\;\;\; -2 \\
\end{align}\]
Now we can add 0 and 4 and put the result in the next position, we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2\text{ }4 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\;1\;\;\; -2\;\;\;4 \\
\end{align}\]
Now we can multiply the newest entry (4) by the divisor (-2) and place the result of (-8) under the next term in the dividend (0), we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2\text{ }4\text{ }-8 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\;1\;\;\;\; -2\;\;\;\ 4\;\;\;\ -8 \\
\end{align}\]
Now we can multiply the newest entry (-8) by the divisor (-2) and place the result of (16) under the next term in the dividend (-16), we get
\[\begin{align}
& -2\text{ }\underline{\left| \begin{align}
& \text{1 0 0 0 }-16 \\
& \downarrow -2\text{ }4\text{ }-8\text{ }16 \\
\end{align} \right.\text{ }} \\
& \;\;\;\;\;\;\;\;\;\;1\;\;\;\; -2\;\;\; 4\;\;\;\; -8\;\;\; 0 \\
\end{align}\]
We know that dividend is equal to quotient plus the remainder over the divisor.
Dividend = Quotient + Remainder/Divisor.
Therefore, the final answer is \[{{x}^{3}}-2{{x}^{2}}+4x-8\].
Note: Students make mistakes while using the synthetic division. We should first differentiate the dividend and the divisor. We should take the first term of the dividend to the result and we should start multiplying it with the divisor to put it for the next value which is added to the next term of the divisor. By repeating this method, we will get the quotient and the remainder value.
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