Question

How do you find the quotient of \[(2{y^2} - 3y + 1) \div (y - 2)\] using long division?

Answer 1

Hint: The long division method for polynomials is used for finding the quotient of the polynomial \[(2{y^2} - 3y + 1)\]by the polynomial \[(y - 2)\]. First, we need to take the first digit of the dividend and need to divide it by the divisor. Then we have to write the answer obtained as a quotient and then multiply this term of the quotient by the divisor to get the result. Then we need to subtract the result from the above dividend and write the difference at the bottom.

Complete answer:
In the given \[(2{y^2} - 3y + 1) \div (y - 2)\], \[(y - 2)\]is the divisor and \[(2{y^2} - 3y + 1)\]is the dividend and we need to find the quotient.
Let us write the given polynomial into a long division form:
\[(y - 2)\left){\vphantom{1{2{y^2} - 3y + 1}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{y^2} - 3y + 1}}}\]
Let us divide \[2{y^2}\]by \[y\]
Now we get \[2y\]which is the quotient. We need to multiply \[2y\]with \[(y - 2)\]
 \[(y - 2)\mathop{\left){\vphantom{1
  2{y^2} - 3y + 1 \\
  2{y^2} - 4y \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  2{y^2} - 3y + 1 \\
  2{y^2} - 4y \\
 }}}
\limits^{\displaystyle \,\,\, {2y}}\]
Now we shall subtract \[2{y^2} - 4y\]from \[2{y^2} - 3y + 1\] bringing \[1\]to bottom
 $\left( {y - 2} \right)\mathop{\left){\vphantom{1
  2{y^2} - 3y + 1 \\
  \underline {2{y^2} - 4y} \\
  0 + y + 1 \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  2{y^2} - 3y + 1 \\
  \underline {2{y^2} - 4y} \\
  0 + y + 1 \\
 }}}
\limits^{\displaystyle \,\,\, {2y}}$
Let us divide \[y\]by \[y\]
Now we get \[1\]which is the quotient. We need to multiply \[1\]with \[(y - 2)\]
$\left( {y - 2} \right)\mathop{\left){\vphantom{1
  2{y^2} - 3y + 1 \\
  \underline {2{y^2} - 4y} \\
  y + 1 \\
  y - 2 \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  2{y^2} - 3y + 1 \\
  \underline {2{y^2} - 4y} \\
  y + 1 \\
  y - 2 \\
 }}}
\limits^{\displaystyle \,\,\, {2y + 1}}$
Now we shall subtract \[y - 2\]from \[y + 1\]
\[\left( {y - 2} \right)\mathop{\left){\vphantom{1
  2{y^2} - 3y + 1 \\
  \underline {2{y^2} - 4y} \\
  y + 1 \\
  \underline {y - 2} \\
  3 \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  2{y^2} - 3y + 1 \\
  \underline {2{y^2} - 4y} \\
  y + 1 \\
  \underline {y - 2} \\
  3 \\
 }}}
\limits^{\displaystyle \,\,\, {2y + 1}}\]
Hence the required quotient of \[(2{y^2} - 3y + 1) \div (y - 2)\]using long division is $2y + 1$ and the remainder of \[(2{y^2} - 3y + 1) \div (y - 2)\]using long division is $3$.

Note:
We need to continue this process of long division until we will get a remainder; and remainder can be zero or of a lower index than that of the divisor. It is very important that while we use the long division method we alter the sign of the terms when subtracting and this small detail of the process may lead to incorrect solutions while computing the long division method of polynomials.