What is the expression obtained when you divide \[{{x}^{3}}-6{{x}^{2}}+11x-6\] by \[{{x}^{2}}+x+1\] ?
Answer
664.8k+ views
Hint: Here we have to use the long division method to divide. The expression obtained will contain the quotient and the remainder.
Complete step by step solution:
In the given question, we have to find the expression for \[\dfrac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}+x+1}\].
So after the long division of the above expression, we get:
\[\Rightarrow \dfrac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}+x+1}=x-7+\dfrac{17x+1}{{{x}^{2}}+x+1}\]
The method for long division is that we need to divide until we get the highest power in the numerator is less than the highest power in the denominator.
Now, the explanation is given as follows.
So here we will first divide\[\dfrac{{{x}^{3}}}{{{x}^{2}}}\], and we will get \[\dfrac{{{x}^{3}}}{{{x}^{2}}}=x\].
Next we will multiply x with \[{{x}^{2}}+x+1\;\]to get \[{{x}^{3}}+{{x}^{2}}+x\].
Then we will subtract \[{{x}^{3}}+{{x}^{2}}+x\] from \[{{x}^{3}}-6{{x}^{2}}+11x-6\] to get the remainder as
\[\begin{align}
& \Rightarrow ({{x}^{3}}-6{{x}^{2}}+11x-6)-({{x}^{3}}+{{x}^{2}}+x) \\
& \Rightarrow -7{{x}^{2}}+10x-6 \\
\end{align}\]
Hence we can write:
\[\Rightarrow \dfrac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}+x+1}=x+\dfrac{-7{{x}^{2}}+10x-6}{{{x}^{2}}+x+1}\]
Next, we will divide\[\dfrac{-7{{x}^{2}}+10x-6}{{{x}^{2}}+x+1}\], so we will first divide \[\dfrac{-7{{x}^{2}}}{{{x}^{2}}}\] to get \[\dfrac{-7{{x}^{2}}}{{{x}^{2}}}=-7\]
Now, we will multiply \[{{x}^{2}}+x+1\;\text{by}\;-7\] to get:
\[\Rightarrow ({{x}^{2}}+x+1)\;\times (\;-7)=-7{{x}^{2}}-7x-7\]
Next, we will subtract \[-7{{x}^{2}}-7x-7\;\text{from}\;-7{{x}^{2}}+10x-6\] to get:
\[\Rightarrow (-7{{x}^{2}}+10x-6)-(-7{{x}^{2}}-7x-7)=\text{17x+1}\]
So, now we can write
\[\Rightarrow \dfrac{-7{{x}^{2}}+10x-6}{{{x}^{2}}+x+1}=-7+\dfrac{17x+1}{{{x}^{2}}+x+1}\]
Now, we can divide further, as the highest power in the numerator is less than the highest power in the denominator.
So, finally we can write:
\[\Rightarrow \dfrac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}+x+1}=x-7+\dfrac{17x+1}{{{x}^{2}}+x+1}\]
Hence the expression obtained after division is \[x-7+\dfrac{17x+1}{{{x}^{2}}+x+1}\].
Note: The division of two algebraic expressions is only possible, when we have the highest power of the expression in the numerator equal or greater than the highest power of the expression in the denominator. The care has to be taken when we are dividing the expression with the other expression, in terms of the power. We need to divide each term in the numerator with the expression in the denominator, for example \[\dfrac{{{x}^{3}}+{{x}^{2}}+x}{x}={{x}^{2}}+{{x}^{1}}+1\]and not \[\dfrac{{{x}^{3}}+{{x}^{2}}+x}{x}\ne {{x}^{2}}+\left( {{x}^{2}}+x \right)\]
Complete step by step solution:
In the given question, we have to find the expression for \[\dfrac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}+x+1}\].
So after the long division of the above expression, we get:
\[\Rightarrow \dfrac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}+x+1}=x-7+\dfrac{17x+1}{{{x}^{2}}+x+1}\]
The method for long division is that we need to divide until we get the highest power in the numerator is less than the highest power in the denominator.
Now, the explanation is given as follows.
So here we will first divide\[\dfrac{{{x}^{3}}}{{{x}^{2}}}\], and we will get \[\dfrac{{{x}^{3}}}{{{x}^{2}}}=x\].
Next we will multiply x with \[{{x}^{2}}+x+1\;\]to get \[{{x}^{3}}+{{x}^{2}}+x\].
Then we will subtract \[{{x}^{3}}+{{x}^{2}}+x\] from \[{{x}^{3}}-6{{x}^{2}}+11x-6\] to get the remainder as
\[\begin{align}
& \Rightarrow ({{x}^{3}}-6{{x}^{2}}+11x-6)-({{x}^{3}}+{{x}^{2}}+x) \\
& \Rightarrow -7{{x}^{2}}+10x-6 \\
\end{align}\]
Hence we can write:
\[\Rightarrow \dfrac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}+x+1}=x+\dfrac{-7{{x}^{2}}+10x-6}{{{x}^{2}}+x+1}\]
Next, we will divide\[\dfrac{-7{{x}^{2}}+10x-6}{{{x}^{2}}+x+1}\], so we will first divide \[\dfrac{-7{{x}^{2}}}{{{x}^{2}}}\] to get \[\dfrac{-7{{x}^{2}}}{{{x}^{2}}}=-7\]
Now, we will multiply \[{{x}^{2}}+x+1\;\text{by}\;-7\] to get:
\[\Rightarrow ({{x}^{2}}+x+1)\;\times (\;-7)=-7{{x}^{2}}-7x-7\]
Next, we will subtract \[-7{{x}^{2}}-7x-7\;\text{from}\;-7{{x}^{2}}+10x-6\] to get:
\[\Rightarrow (-7{{x}^{2}}+10x-6)-(-7{{x}^{2}}-7x-7)=\text{17x+1}\]
So, now we can write
\[\Rightarrow \dfrac{-7{{x}^{2}}+10x-6}{{{x}^{2}}+x+1}=-7+\dfrac{17x+1}{{{x}^{2}}+x+1}\]
Now, we can divide further, as the highest power in the numerator is less than the highest power in the denominator.
So, finally we can write:
\[\Rightarrow \dfrac{{{x}^{3}}-6{{x}^{2}}+11x-6}{{{x}^{2}}+x+1}=x-7+\dfrac{17x+1}{{{x}^{2}}+x+1}\]
Hence the expression obtained after division is \[x-7+\dfrac{17x+1}{{{x}^{2}}+x+1}\].
Note: The division of two algebraic expressions is only possible, when we have the highest power of the expression in the numerator equal or greater than the highest power of the expression in the denominator. The care has to be taken when we are dividing the expression with the other expression, in terms of the power. We need to divide each term in the numerator with the expression in the denominator, for example \[\dfrac{{{x}^{3}}+{{x}^{2}}+x}{x}={{x}^{2}}+{{x}^{1}}+1\]and not \[\dfrac{{{x}^{3}}+{{x}^{2}}+x}{x}\ne {{x}^{2}}+\left( {{x}^{2}}+x \right)\]
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