Question

Use synthetic division method for performing the following division. Write the results in the form:
Dividend \[ = \] Divisor \[ \times \] Quotient \[ + \] Remainder
\[\left( {{y^2} - 11y + 30} \right) \div \left( {y - 5} \right)\]
A.\[\left( {y - 7} \right)\left( {y + 1} \right) + 0\]
B.\[\left( {y + 4} \right)\left( {y + 6} \right) + 0\]
C.\[\left( y \right)\left( {y - 5} \right) + 0\]
D.\[\left( {y - 5} \right)\left( {y - 6} \right) + 0\]

Answer 1

Hint: Here, we are required to find the quotient and remainder using the synthetic division method. Here, we will only write the coefficients and then follow the further process . We will solve this question using this method and find the quotient and the remainder. Finally, we will express our answer in the form of Dividend \[ = \] Divisor \[ \times \] Quotient\[ + \] Remainder and hence, reach the required answer.

Complete step-by-step answer:
The synthetic division method is a shortcut for dividing two algebraic expressions.
In this method, we write only the coefficients instead of the complete dividend.
Here, Dividend \[ = {y^2} - 11y + 30\]
Hence the coefficients are 1, \[ - 11\] and 30.
Also, the Divisor\[ = y - 5\]
For this, we will take \[ + 5\] as the divisor.
We will use the following process to divide the given number using the synthetic division method:
Step 1: We will write the coefficients as shown.
Step 2: Then we will take down the first coefficient without changing it.
Step 3: we will then multiply it by the divisor, take the number under the second coefficient, and then add them.
Step 4: again, we will multiply the sum under the second coefficient by the divisor, take that number under the third coefficient, and then add them.
Step 5: Finally, we will convert the first two coefficients as the quotient and the third as the remainder and this is the required answer.
\[\begin{array}{l}5\left| \!{\underline {\, {\begin{array}{*{20}{l}}1&{ - 11}&{30}\\ \downarrow &5&{ - 30}\end{array}} \,}} \right. \\\begin{array}{*{20}{c}} \times &1&{ - 6}&0\end{array}\end{array}\]
Now, the first two coefficients show the quotient and since our dividend was a quadratic equation, the quotient will be a linear equation. Hence, \[1, - 6\] will be written as: \[y - 6\] and the remainder is 0.
Therefore, if we write the results in the form:
Dividend \[ = \] Divisor \[ \times \] Quotient \[ + \] Remainder
\[ \Rightarrow {y^2} - 11y + 30 = \left( {y - 5} \right)\left( {y - 6} \right) + 0\]
Hence, option D is the correct answer.

Note: Another way of attempting this question is:
According to the question,
Dividend \[ = {y^2} - 11y + 30\]
Divisor \[ = y - 5\]
Now, doing middle term split of the dividend,
\[{y^2} - 11y + 30 = {y^2} - 6y - 5y + 30\]
This can be written as:
\[ \Rightarrow {y^2} - 11y + 30 = y\left( {y - 6} \right) - 5\left( {y - 6} \right)\]
Factoring out the common terms, we get
\[ \Rightarrow {y^2} - 11y + 30 = \left( {y - 5} \right)\left( {y - 6} \right)\]
Now, we will divide this dividend by the given divisor. Therefore, we get
\[ \Rightarrow {y^2} - 11y + 30 = \dfrac{{\left( {y - 5} \right)\left( {y - 6} \right)}}{{\left( {y - 5} \right)}} = \left( {y - 6} \right)\]
Hence, quotient \[ = \left( {y - 6} \right)\] and remainder \[ = 0\]
Therefore, if we write the results in the form:
Dividend \[ = \] Divisor \[ \times \] Quotient \[ + \] Remainder
\[ \Rightarrow {y^2} - 11y + 30 = \left( {y - 5} \right)\left( {y - 6} \right) + 0\]
Hence, option D is the correct answer.