Question 189558
The last post is correct (even though weird notation was used), but s/he didn't use synthetic division.




Note: the test zero is x=-3



First set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.

<TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>4</TD><TD>-3</TD><TD>-5</TD><TD>9</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>



Now bring down the leading coefficient (it is the coefficient with the highest exponent which is 4)

<TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>4</TD><TD>-3</TD><TD>-5</TD><TD>9</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -3 by 4 and place the product (which is -12)  right underneath the second  coefficient (which is -3)

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>4</TD><TD>-3</TD><TD>-5</TD><TD>9</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-12</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -12 and -3 to get -15. Place the sum right underneath -12.

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>4</TD><TD>-3</TD><TD>-5</TD><TD>9</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-12</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>-15</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -3 by -15 and place the product (which is 45)  right underneath the third  coefficient (which is -5)

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>4</TD><TD>-3</TD><TD>-5</TD><TD>9</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-12</TD><TD>45</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>-15</TD><TD></TD><TD></TD></TR></TABLE>

    Add 45 and -5 to get 40. Place the sum right underneath 45.

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>4</TD><TD>-3</TD><TD>-5</TD><TD>9</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-12</TD><TD>45</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>-15</TD><TD>40</TD><TD></TD></TR></TABLE>

    Multiply -3 by 40 and place the product (which is -120)  right underneath the fourth  coefficient (which is 9)

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>4</TD><TD>-3</TD><TD>-5</TD><TD>9</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-12</TD><TD>45</TD><TD>-120</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>-15</TD><TD>40</TD><TD></TD></TR></TABLE>

    Add -120 and 9 to get -111. Place the sum right underneath -120.

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>4</TD><TD>-3</TD><TD>-5</TD><TD>9</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-12</TD><TD>45</TD><TD>-120</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>-15</TD><TD>40</TD><TD>-111</TD></TR></TABLE>


Since the last column adds to -111, we have a remainder of -111. 



So according to the remainder theorem, this means that {{{f(-3)=-111}}} which tells us that the answer is A.