Question

Divide the polynomial \[{\rm{3}}{{\rm{x}}^{\rm{4}}} - {\rm{4}}{{\rm{x}}^{\rm{3}}} - {\rm{3x}} - {\rm{1}}\] by \[x - 1\].

Answer 1

Hint: The division questions are solved by dividing only the first variable of dividend by the first variable of divisor, subtracting the remaining part, then again dividing the first part of the remaining dividend by the first part of divisor, and repeating this process or methodology over and over again till we reach a point where the remainder is a constant, which includes zero. We can also verify the relation by putting in the values of dividend, divisor, quotient and remainder into the Euclid’s division lemma and checking the two sides of equality.

Formula Used:
We are going to use the Euclid’s division lemma to verify the relation that we obtain, which is:
dividend \[ = \] divisor\[ \times \]quotient \[ + \] remainder
or, in mathematical form it is represented as:
\[a = bq + r \;\;\;\;\;\;\;\;\;\] \[0 \le r < b\]
where \[a\] is the dividend, \[b\] is the divisor, \[q\] is the quotient and \[r\] is the remainder.

Complete step by step answer:
First, we divide \[3{x^4}\] (first term of the dividend) by \[x\] (first term of the divisor)
\[\dfrac{{3{x^4}}}{x} = 3{x^3}\]
So, that means that the first term of the quotient is \[3{x^3}\].
\[x - 1\mathop{\left){\vphantom{1\begin{array}{l}3{x^4} - 4{x^3} - 3x - 1\\3{x^4} - 3{x^3}\\\dfrac{{ - {\rm{ }} + {\rm{ }}}}{{0 - {x^3}}}\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}3{x^4} - 4{x^3} - 3x - 1\\3{x^4} - 3{x^3}\\\dfrac{{ - {\rm{ }} + {\rm{ }}}}{{0 - {x^3}}}\end{array}}}}
\limits^{\displaystyle\,\,\, {3{x^3}}}\]

Now, we divide the first term of remaining divisor by \[x\]
\[\dfrac{{ - {x^3}}}{x} = - {x^2}\]
\[x - 1\mathop{\left){\vphantom{1\begin{array}{l}3{x^4} - 4{x^3} - 3x - 1\\3{x^4} - 3{x^3}\\\dfrac{{ - {\rm{ }} + {\rm{ }}}}{\begin{array}{l}0 - {x^3}\\{\rm{ }}{x^3} - 3x\\{\rm{ }}{x^3} + {x^2}\\\dfrac{{ - {\rm{ }} - {\rm{ }}}}{{0 - {x^2} - 3x - 1}}{\rm{ }}\end{array}}\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}3{x^4} - 4{x^3} - 3x - 1\\3{x^4} - 3{x^3}\\\dfrac{{ - {\rm{ }} + {\rm{ }}}}{\begin{array}{l}0 - {x^3}\\{\rm{ }}{x^3} - 3x\\{\rm{ }}{x^3} + {x^2}\\\dfrac{{ - {\rm{ }} - {\rm{ }}}}{{0 - {x^2} - 3x - 1}}{\rm{ }}\end{array}}\end{array}}}}
\limits^{\displaystyle\,\,\, {3{x^3} - {x^2}}}\]
Now, we divide \[ - {x^2}\] by \[x\]
\[\dfrac{{ - {x^2}}}{x} = - x\]
\[x - 1\mathop{\left){\vphantom{1\begin{array}{l}3{x^4} - 4{x^3} - 3x - 1\\3{x^4} - 3{x^3}\\\dfrac{{ - {\rm{ }} + {\rm{ }}}}{\begin{array}{l}0 - {x^3}\\{\rm{ }}{x^3} - 3x\\{\rm{ }}{x^3} + {x^2}\\\dfrac{{ - {\rm{ }} - {\rm{ }}}}{\begin{array}{l}0 - {x^2} - 3x - 1\\{\rm{ }} - {x^2} + x\\\dfrac{{ + {\rm{ }} - }}{{0 - 4x - 1}}\end{array}}{\rm{ }}\end{array}}\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}3{x^4} - 4{x^3} - 3x - 1\\3{x^4} - 3{x^3}\\\dfrac{{ - {\rm{ }} + {\rm{ }}}}{\begin{array}{l}0 - {x^3}\\{\rm{ }}{x^3} - 3x\\{\rm{ }}{x^3} + {x^2}\\\dfrac{{ - {\rm{ }} - {\rm{ }}}}{\begin{array}{l}0 - {x^2} - 3x - 1\\{\rm{ }} - {x^2} + x\\\dfrac{{ + {\rm{ }} - }}{{0 - 4x - 1}}\end{array}}{\rm{ }}\end{array}}\end{array}}}}
\limits^{\displaystyle\,\,\, {3{x^3} - {x^2} - x}}\]
Finally, we divide \[ - 4x\] by \[x\]
\[\dfrac{{ - 4x}}{x} = - 4\]
\[x - 1\mathop{\left){\vphantom{1\begin{array}{l}3{x^4} - 4{x^3} - 3x - 1\\3{x^4} - 3{x^3}\\\dfrac{{ - {\rm{ }} + {\rm{ }}}}{\begin{array}{l}0 - {x^3}\\{\rm{ }}{x^3} - 3x\\{\rm{ }}{x^3} + {x^2}\\\dfrac{{ - {\rm{ }} - {\rm{ }}}}{\begin{array}{l}0 - {x^2} - 3x - 1\\{\rm{ }} - {x^2} + x\\\dfrac{{ + {\rm{ }} - }}{\begin{array}{l}0 - 4x - 1\\{\rm{ }} - 4x + 4\\{\rm{ }}\dfrac{{{\rm{ }} + {\rm{ }} - }}{{{\rm{ }}\dfrac{{{\rm{ }} - 5{\rm{ }}}}{{}}}}\end{array}}\end{array}}{\rm{ }}\end{array}}\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}3{x^4} - 4{x^3} - 3x - 1\\3{x^4} - 3{x^3}\\\dfrac{{ - {\rm{ }} + {\rm{ }}}}{\begin{array}{l}0 - {x^3}\\{\rm{ }}{x^3} - 3x\\{\rm{ }}{x^3} + {x^2}\\\dfrac{{ - {\rm{ }} - {\rm{ }}}}{\begin{array}{l}0 - {x^2} - 3x - 1\\{\rm{ }} - {x^2} + x\\\dfrac{{ + {\rm{ }} - }}{\begin{array}{l}0 - 4x - 1\\{\rm{ }} - 4x + 4\\{\rm{ }}\dfrac{{{\rm{ }} + {\rm{ }} - }}{{{\rm{ }}\dfrac{{{\rm{ }} - 5{\rm{ }}}}{{}}}}\end{array}}\end{array}}{\rm{ }}\end{array}}\end{array}}}}
\limits^{\displaystyle\,\,\, {3{x^3} - {x^2} - x - 4}}\]
Thus, on dividing \[{\rm{3}}{{\rm{x}}^{\rm{4}}} - {\rm{4}}{{\rm{x}}^{\rm{3}}} - {\rm{3x}} - {\rm{1}}\] by \[x - 1\], we get \[3{x^3} - {x^2} - x - 4\] as quotient and \[5\] as the remainder.

Note: So, in this question we saw that solving this type of question is very easy. All one needs to know is how to methodically approach the question, follow the steps religiously, checking each step twice (in case one misses something, as going back to look for the error after the final step has been reached is very time consuming and exhilarating), verifying the answer with the Euclid’s division lemma by putting in the values of dividend, divisor, quotient and remainder into it and comparing the two sides of the equality.