Question 134154
To find f(4), we can use synthetic division with the test zero of 4



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>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><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><TD></TD></TR></TABLE>

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 6)

<TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 4 by 6 and place the product (which is 24)  right underneath the second  coefficient (which is 11)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>24</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 24 and 11 to get 35. Place the sum right underneath 24.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>24</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>35</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 4 by 35 and place the product (which is 140)  right underneath the third  coefficient (which is 6)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>24</TD><TD>140</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>35</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 140 and 6 to get 146. Place the sum right underneath 140.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>24</TD><TD>140</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>35</TD><TD>146</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 4 by 146 and place the product (which is 584)  right underneath the fourth  coefficient (which is -6)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>24</TD><TD>140</TD><TD>584</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>35</TD><TD>146</TD><TD></TD><TD></TD></TR></TABLE>

    Add 584 and -6 to get 578. Place the sum right underneath 584.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>24</TD><TD>140</TD><TD>584</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>35</TD><TD>146</TD><TD>578</TD><TD></TD></TR></TABLE>

    Multiply 4 by 578 and place the product (which is 2312)  right underneath the fifth  coefficient (which is 53)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>24</TD><TD>140</TD><TD>584</TD><TD>2312</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>35</TD><TD>146</TD><TD>578</TD><TD></TD></TR></TABLE>

    Add 2312 and 53 to get 2365. Place the sum right underneath 2312.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>6</TD><TD>11</TD><TD>6</TD><TD>-6</TD><TD>53</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>24</TD><TD>140</TD><TD>584</TD><TD>2312</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>35</TD><TD>146</TD><TD>578</TD><TD>2365</TD></TR></TABLE>

Since the last column adds to 2365, we have a remainder of 2365. 



So {{{f(4)=2365}}}