Divide the polynomial \[p\left( x \right)\] by the polynomial \[g\left( x \right)\] and find the quotient and remainder in each of the following:
i). \[p\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x-3\], \[g\left( x \right)={{x}^{2}}-2\]
ii). \[p\left( x \right)={{x}^{4}}-3{{x}^{2}}+4x+5\], \[g\left( x \right)={{x}^{2}}-x+1\]
iii). \[p\left( x \right)={{x}^{3}}-5x+6\], \[g\left( x \right)=2-{{x}^{2}}\]
Answer
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Hint: In order to divide the polynomial \[p\left( x \right)\] by the polynomial \[g\left( x \right)\] and find the quotient and remainder in each of the given cases, use long division polynomial method that is follows: Divide the first term of the numerator by the first term of the denominator, and put that in the answer. Then multiply the denominator by that answer, put that below the numerator. Then subtract to create a new polynomial repeat it until you can’t divide it further.
Complete step by step solution:
According to the question, let’s divide the polynomial \[p\left( x \right)\] by the polynomial \[g\left( x \right)\] and find the quotient and remainder in each of the following:
i). \[p\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x-3\], \[g\left( x \right)={{x}^{2}}-2\]
Write the problem in the division format:
\[{{x}^{2}}-2\overline{\left){{{x}^{3}}-3{{x}^{2}}+5x-3}\right.}\]
Apply long division method:
Step 1: Divide the leading term of the dividend by the leading term of the divisor: \[\dfrac{{{x}^{3}}}{{{x}^{2}}}=x\] then write down the calculated result in the upper part of the table.
Multiply it by the divisor: \[x\left( {{x}^{2}}-2 \right)={{x}^{3}}-2x\]. Subtract the obtained result from the dividend: \[\left( {{x}^{3}}-3{{x}^{2}}+5x-3 \right)-\left( {{x}^{3}}-2x \right)=-3{{x}^{2}}+7x-3\].
Hence, we can write this step as follows:
\[{{x}^{2}}-2\overset{x}{\overline{\left){\begin{align}
& \text{ }{{x}^{3}}-3{{x}^{2}}+5x-3 \\
& \underline{-{{x}^{3}}+0{{x}^{2}}+2x\text{ }} \\
& \text{ }-3{{x}^{2}}+7x-3 \\
\end{align}}\right.}}\]
Step 2: Divide the leading term of the obtained remainder by the leading term of the divisor: \[\dfrac{-3{{x}^{2}}}{{{x}^{2}}}=-3\] then write down the calculated result in the upper part of the table.
Multiply it by the divisor: \[-3\left( {{x}^{2}}-2 \right)=-3{{x}^{2}}+6\] and subtract the remainder from the obtained result: \[\left( -3{{x}^{2}}+7x-3 \right)-\left( -3{{x}^{2}}+6 \right)=7x-9\].
Hence, we can write this step as follows:
\[{{x}^{2}}-2\overset{x-3}{\overline{\left){\begin{align}
& \text{ }{{x}^{3}}-3{{x}^{2}}+5x-3 \\
& \underline{-{{x}^{3}}+0{{x}^{2}}+2x\text{ }} \\
& \text{ }-3{{x}^{2}}+7x-3 \\
& \text{ }\underline{\text{ }+3{{x}^{2}}+0x-6} \\
& \text{ }7x-9 \\
\end{align}}\right.}}\]
Since the degree of the remainder is less than the degree of the divisor, then we are done.
Therefore, \[\dfrac{{{x}^{3}}-3{{x}^{2}}+5x-3}{{{x}^{2}}-2}=x-3+\dfrac{7x-9}{{{x}^{2}}-2}\].
Hence the quotient is \[x-3\] and the remainder is \[7x-9\].
ii). \[p\left( x \right)={{x}^{4}}-3{{x}^{2}}+4x+5\], \[g\left( x \right)={{x}^{2}}-x+1\]
Write the problem in the special format (missed terms are written with zero coefficients):
\[{{x}^{2}}-x+1\overline{\left){{{x}^{4}}+0{{x}^{3}}-3{{x}^{2}}+4x+5}\right.}\]
Step 1: Divide the leading term of the dividend by the leading term of the divisor: \[\dfrac{{{x}^{4}}}{{{x}^{2}}}={{x}^{2}}\]then
write down the calculated result in the upper part of the table. Multiply it by the divisor: \[{{x}^{2}}\left( {{x}^{2}}-x+1 \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}}\] and subtract the dividend from the obtained result: \[\left( {{x}^{4}}-3{{x}^{2}}+4x+5 \right)-\left( {{x}^{4}}-{{x}^{3}}+{{x}^{2}} \right)={{x}^{3}}-4{{x}^{2}}+4x+5\]
Hence, we can write this step as follows:
\[{{x}^{2}}-x+1\overset{{{x}^{2}}}{\overline{\left){\begin{align}
& \text{ }{{x}^{4}}+0{{x}^{3}}-3{{x}^{2}}+4x+5 \\
& \underline{-{{x}^{4}}+{{x}^{3}}\text{ }-{{x}^{2}}\text{ }} \\
& \text{ }{{x}^{3}}-4{{x}^{2}}+4x+5 \\
\end{align}}\right.}}\]
Step 2: Divide the leading term of the obtained remainder by the leading term of the divisor: \[\dfrac{{{x}^{3}}}{{{x}^{2}}}=x\]then write down the calculated result in the upper part of the table. Multiply it by the divisor: \[x\left( {{x}^{2}}-x+1 \right)={{x}^{3}}-{{x}^{2}}+x\]and subtract the remainder from the obtained result: \[\left( {{x}^{3}}-4{{x}^{2}}+4x+5 \right)-\left( {{x}^{3}}-{{x}^{2}}+x \right)=-3{{x}^{2}}+3x+5\]
Hence, we can write this step as follows:
\[{{x}^{2}}-x+1\overset{{{x}^{2}}+x}{\overline{\left){\begin{align}
& \text{ }{{x}^{4}}+0{{x}^{3}}-3{{x}^{2}}+4x+5 \\
& \underline{-{{x}^{4}}-{{x}^{3}}\text{ }-{{x}^{2}}\text{ }} \\
& \text{ }{{x}^{3}}-4{{x}^{2}}+4x+5 \\
& \text{ }\underline{\text{ }-{{x}^{3}}\text{+ }{{x}^{2}}-\text{ }x\text{ }} \\
& \text{ }-3{{x}^{2}}+3x+5\text{ } \\
\end{align}}\right.}}\]
Step 3: Divide the leading term of the obtained remainder by the leading term of the divisor: \[\dfrac{-3{{x}^{2}}}{{{x}^{2}}}=-3\] then write down the calculated result in the upper part of the table.
Multiply it by the divisor: \[-3\left( {{x}^{2}}-x+1 \right)=-3{{x}^{2}}+3x-3\] and subtract the remainder from the obtained result: \[\left( -3{{x}^{2}}+3x+5 \right)-\left( -3{{x}^{2}}+3x-3 \right)=8\]
Hence, we can write this step as follows:
\[{{x}^{2}}-x+1\overset{{{x}^{2}}+x-3}{\overline{\left){\begin{align}
& \text{ }{{x}^{4}}+0{{x}^{3}}-3{{x}^{2}}+4x+5 \\
& \underline{-{{x}^{4}}-{{x}^{3}}\text{ }-{{x}^{2}}\text{ }} \\
& \text{ }{{x}^{3}}-4{{x}^{2}}+4x+5 \\
& \text{ }\underline{\text{ }-{{x}^{3}}\text{+ }{{x}^{2}}-\text{ }x\text{ }} \\
& \text{ }-3{{x}^{2}}+3x+5 \\
& \text{ }\underline{\text{ }+3{{x}^{2}}-3x+3} \\
& \text{ }8\text{ } \\
\end{align}}\right.}}\]
Since the degree of the remainder is less than the degree of the divisor, then we are done.
Therefore, \[\dfrac{{{x}^{4}}-3{{x}^{2}}+4x+5}{{{x}^{2}}-x+1}={{x}^{2}}+x-3+\dfrac{8}{{{x}^{2}}-x+1}\].
Hence the quotient is \[{{x}^{2}}+x-3\] and the remainder is \[8\].
iii). \[p\left( x \right)={{x}^{3}}-5x+6\], \[g\left( x \right)=2-{{x}^{2}}\]
Write the problem in the special format (missed terms are written with zero coefficients):
\[-{{x}^{2}}+2\overline{\left){{{x}^{3}}+0{{x}^{2}}-5x+6}\right.}\]
Step 1: Divide the leading term of the dividend by the leading term of the divisor: \[\dfrac{{{x}^{3}}}{-{{x}^{2}}}=-x\] then
write down the calculated result in the upper part of the table. Multiply it by the divisor: \[-x\left( -{{x}^{2}}+2 \right)={{x}^{3}}-2x\] and subtract the dividend from the obtained result: \[\left( {{x}^{3}}-5x+6 \right)-\left( {{x}^{3}}-2x \right)=-3x+6\]
Hence, we can write this step as follows:
\[-{{x}^{2}}+2\overset{-x}{\overline{\left){\begin{align}
& \text{ }{{x}^{3}}+0{{x}^{2}}-5x+6 \\
& \underline{-{{x}^{3}}+0{{x}^{2}}+2x\text{ }} \\
& \text{ }-3x+6 \\
\end{align}}\right.}}\]
Since the degree of the remainder is less than the degree of the divisor, then we are done.
Therefore, \[\dfrac{{{x}^{3}}-5x+6}{-{{x}^{2}}+2}=-x+\dfrac{6-3x}{2-{{x}^{2}}}\].
Hence the quotient is \[-x\] and the remainder is \[-3x+6\].
Note: Key point is to remember the concept of long division polynomials where long division is follows: divide the first term of the numerator by the first term of the denominator, and put that in the answer. Then multiply the denominator by that answer, put that below the numerator. Then subtract to create a new polynomial repeat it until you can’t divide it further.
Complete step by step solution:
According to the question, let’s divide the polynomial \[p\left( x \right)\] by the polynomial \[g\left( x \right)\] and find the quotient and remainder in each of the following:
i). \[p\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x-3\], \[g\left( x \right)={{x}^{2}}-2\]
Write the problem in the division format:
\[{{x}^{2}}-2\overline{\left){{{x}^{3}}-3{{x}^{2}}+5x-3}\right.}\]
Apply long division method:
Step 1: Divide the leading term of the dividend by the leading term of the divisor: \[\dfrac{{{x}^{3}}}{{{x}^{2}}}=x\] then write down the calculated result in the upper part of the table.
Multiply it by the divisor: \[x\left( {{x}^{2}}-2 \right)={{x}^{3}}-2x\]. Subtract the obtained result from the dividend: \[\left( {{x}^{3}}-3{{x}^{2}}+5x-3 \right)-\left( {{x}^{3}}-2x \right)=-3{{x}^{2}}+7x-3\].
Hence, we can write this step as follows:
\[{{x}^{2}}-2\overset{x}{\overline{\left){\begin{align}
& \text{ }{{x}^{3}}-3{{x}^{2}}+5x-3 \\
& \underline{-{{x}^{3}}+0{{x}^{2}}+2x\text{ }} \\
& \text{ }-3{{x}^{2}}+7x-3 \\
\end{align}}\right.}}\]
Step 2: Divide the leading term of the obtained remainder by the leading term of the divisor: \[\dfrac{-3{{x}^{2}}}{{{x}^{2}}}=-3\] then write down the calculated result in the upper part of the table.
Multiply it by the divisor: \[-3\left( {{x}^{2}}-2 \right)=-3{{x}^{2}}+6\] and subtract the remainder from the obtained result: \[\left( -3{{x}^{2}}+7x-3 \right)-\left( -3{{x}^{2}}+6 \right)=7x-9\].
Hence, we can write this step as follows:
\[{{x}^{2}}-2\overset{x-3}{\overline{\left){\begin{align}
& \text{ }{{x}^{3}}-3{{x}^{2}}+5x-3 \\
& \underline{-{{x}^{3}}+0{{x}^{2}}+2x\text{ }} \\
& \text{ }-3{{x}^{2}}+7x-3 \\
& \text{ }\underline{\text{ }+3{{x}^{2}}+0x-6} \\
& \text{ }7x-9 \\
\end{align}}\right.}}\]
Since the degree of the remainder is less than the degree of the divisor, then we are done.
Therefore, \[\dfrac{{{x}^{3}}-3{{x}^{2}}+5x-3}{{{x}^{2}}-2}=x-3+\dfrac{7x-9}{{{x}^{2}}-2}\].
Hence the quotient is \[x-3\] and the remainder is \[7x-9\].
ii). \[p\left( x \right)={{x}^{4}}-3{{x}^{2}}+4x+5\], \[g\left( x \right)={{x}^{2}}-x+1\]
Write the problem in the special format (missed terms are written with zero coefficients):
\[{{x}^{2}}-x+1\overline{\left){{{x}^{4}}+0{{x}^{3}}-3{{x}^{2}}+4x+5}\right.}\]
Step 1: Divide the leading term of the dividend by the leading term of the divisor: \[\dfrac{{{x}^{4}}}{{{x}^{2}}}={{x}^{2}}\]then
write down the calculated result in the upper part of the table. Multiply it by the divisor: \[{{x}^{2}}\left( {{x}^{2}}-x+1 \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}}\] and subtract the dividend from the obtained result: \[\left( {{x}^{4}}-3{{x}^{2}}+4x+5 \right)-\left( {{x}^{4}}-{{x}^{3}}+{{x}^{2}} \right)={{x}^{3}}-4{{x}^{2}}+4x+5\]
Hence, we can write this step as follows:
\[{{x}^{2}}-x+1\overset{{{x}^{2}}}{\overline{\left){\begin{align}
& \text{ }{{x}^{4}}+0{{x}^{3}}-3{{x}^{2}}+4x+5 \\
& \underline{-{{x}^{4}}+{{x}^{3}}\text{ }-{{x}^{2}}\text{ }} \\
& \text{ }{{x}^{3}}-4{{x}^{2}}+4x+5 \\
\end{align}}\right.}}\]
Step 2: Divide the leading term of the obtained remainder by the leading term of the divisor: \[\dfrac{{{x}^{3}}}{{{x}^{2}}}=x\]then write down the calculated result in the upper part of the table. Multiply it by the divisor: \[x\left( {{x}^{2}}-x+1 \right)={{x}^{3}}-{{x}^{2}}+x\]and subtract the remainder from the obtained result: \[\left( {{x}^{3}}-4{{x}^{2}}+4x+5 \right)-\left( {{x}^{3}}-{{x}^{2}}+x \right)=-3{{x}^{2}}+3x+5\]
Hence, we can write this step as follows:
\[{{x}^{2}}-x+1\overset{{{x}^{2}}+x}{\overline{\left){\begin{align}
& \text{ }{{x}^{4}}+0{{x}^{3}}-3{{x}^{2}}+4x+5 \\
& \underline{-{{x}^{4}}-{{x}^{3}}\text{ }-{{x}^{2}}\text{ }} \\
& \text{ }{{x}^{3}}-4{{x}^{2}}+4x+5 \\
& \text{ }\underline{\text{ }-{{x}^{3}}\text{+ }{{x}^{2}}-\text{ }x\text{ }} \\
& \text{ }-3{{x}^{2}}+3x+5\text{ } \\
\end{align}}\right.}}\]
Step 3: Divide the leading term of the obtained remainder by the leading term of the divisor: \[\dfrac{-3{{x}^{2}}}{{{x}^{2}}}=-3\] then write down the calculated result in the upper part of the table.
Multiply it by the divisor: \[-3\left( {{x}^{2}}-x+1 \right)=-3{{x}^{2}}+3x-3\] and subtract the remainder from the obtained result: \[\left( -3{{x}^{2}}+3x+5 \right)-\left( -3{{x}^{2}}+3x-3 \right)=8\]
Hence, we can write this step as follows:
\[{{x}^{2}}-x+1\overset{{{x}^{2}}+x-3}{\overline{\left){\begin{align}
& \text{ }{{x}^{4}}+0{{x}^{3}}-3{{x}^{2}}+4x+5 \\
& \underline{-{{x}^{4}}-{{x}^{3}}\text{ }-{{x}^{2}}\text{ }} \\
& \text{ }{{x}^{3}}-4{{x}^{2}}+4x+5 \\
& \text{ }\underline{\text{ }-{{x}^{3}}\text{+ }{{x}^{2}}-\text{ }x\text{ }} \\
& \text{ }-3{{x}^{2}}+3x+5 \\
& \text{ }\underline{\text{ }+3{{x}^{2}}-3x+3} \\
& \text{ }8\text{ } \\
\end{align}}\right.}}\]
Since the degree of the remainder is less than the degree of the divisor, then we are done.
Therefore, \[\dfrac{{{x}^{4}}-3{{x}^{2}}+4x+5}{{{x}^{2}}-x+1}={{x}^{2}}+x-3+\dfrac{8}{{{x}^{2}}-x+1}\].
Hence the quotient is \[{{x}^{2}}+x-3\] and the remainder is \[8\].
iii). \[p\left( x \right)={{x}^{3}}-5x+6\], \[g\left( x \right)=2-{{x}^{2}}\]
Write the problem in the special format (missed terms are written with zero coefficients):
\[-{{x}^{2}}+2\overline{\left){{{x}^{3}}+0{{x}^{2}}-5x+6}\right.}\]
Step 1: Divide the leading term of the dividend by the leading term of the divisor: \[\dfrac{{{x}^{3}}}{-{{x}^{2}}}=-x\] then
write down the calculated result in the upper part of the table. Multiply it by the divisor: \[-x\left( -{{x}^{2}}+2 \right)={{x}^{3}}-2x\] and subtract the dividend from the obtained result: \[\left( {{x}^{3}}-5x+6 \right)-\left( {{x}^{3}}-2x \right)=-3x+6\]
Hence, we can write this step as follows:
\[-{{x}^{2}}+2\overset{-x}{\overline{\left){\begin{align}
& \text{ }{{x}^{3}}+0{{x}^{2}}-5x+6 \\
& \underline{-{{x}^{3}}+0{{x}^{2}}+2x\text{ }} \\
& \text{ }-3x+6 \\
\end{align}}\right.}}\]
Since the degree of the remainder is less than the degree of the divisor, then we are done.
Therefore, \[\dfrac{{{x}^{3}}-5x+6}{-{{x}^{2}}+2}=-x+\dfrac{6-3x}{2-{{x}^{2}}}\].
Hence the quotient is \[-x\] and the remainder is \[-3x+6\].
Note: Key point is to remember the concept of long division polynomials where long division is follows: divide the first term of the numerator by the first term of the denominator, and put that in the answer. Then multiply the denominator by that answer, put that below the numerator. Then subtract to create a new polynomial repeat it until you can’t divide it further.
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