Question

Find the quotient when the polynomial $12x^3+9x^2+11$ is divided by $x-9$ using synthetic division method.

• A. $22x^2+117x+1053$
• B. $12x^2+137x+1053$
• C. $12x^2+117x+1053$
• D. $12x^2+117x+1057$

Answer 1

The correct option is C $12x^2+117x+1053$

Index form of the dividend polynomial is :

$12x^3+9x^2+11=12x^3+9x^2+0x+11$

Coefficient form of the dividend polynomial is :.

(12,9,0,11)

Divisor polynomial =x-9

Steps:
1. Draw the horizontal and vertical lines as below :



2. Divisor is x-9. Take opposite number of -9 which is 9.Write 9 to the left of the vertical line. Write the coefficient form of the dividend polynomial in the first row and write the first coefficient as it is in the third row

$\begin{array}{ccccc}& x^3& x^2& x^1& x^0\\ & 12& 9& 0& 11\\ 9& \downarrow & & & \\ & 12& & & \end{array}$

3. The product of 12 in the third row
with divisor 9 is 108. Write this 108 in the second row below the coefficient 9. Addition of 9 and 108 which is 117, is to be written in the third row.

Similarly by multiplying and adding, we have

$\begin{array}{ccccc}& x^3& x^2& x^1& x^0\\ & 12& 9& 0& 11\\ 9& \downarrow & 108& 1053& 9477\\ & 12& 117& 1053& |-9488\end{array}$

The last addition is the remainder, which is ' 9488 '.

The coefficient form of the quotient is (12, 117, 1053).

So, quotient is $12x^2+117x+1053$, remainder is '9488'