Question 147505
Since we want to find f(3), this means that our test zero is 3


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the function to the right of the test zero.<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</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 3)

<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</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>3</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 3 by 3 and place the product (which is 9)  right underneath the second  coefficient (which is 8)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>9</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>3</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 9 and 8 to get 17. Place the sum right underneath 9.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>9</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>3</TD><TD>17</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 3 by 17 and place the product (which is 51)  right underneath the third  coefficient (which is 2)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>9</TD><TD>51</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>3</TD><TD>17</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 51 and 2 to get 53. Place the sum right underneath 51.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>9</TD><TD>51</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>3</TD><TD>17</TD><TD>53</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 3 by 53 and place the product (which is 159)  right underneath the fourth  coefficient (which is -7)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>9</TD><TD>51</TD><TD>159</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>3</TD><TD>17</TD><TD>53</TD><TD></TD><TD></TD></TR></TABLE>

    Add 159 and -7 to get 152. Place the sum right underneath 159.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>9</TD><TD>51</TD><TD>159</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>3</TD><TD>17</TD><TD>53</TD><TD>152</TD><TD></TD></TR></TABLE>

    Multiply 3 by 152 and place the product (which is 456)  right underneath the fifth  coefficient (which is -4)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>9</TD><TD>51</TD><TD>159</TD><TD>456</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>3</TD><TD>17</TD><TD>53</TD><TD>152</TD><TD></TD></TR></TABLE>

    Add 456 and -4 to get 452. Place the sum right underneath 456.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>3</TD><TD>8</TD><TD>2</TD><TD>-7</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>9</TD><TD>51</TD><TD>159</TD><TD>456</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>3</TD><TD>17</TD><TD>53</TD><TD>152</TD><TD>452</TD></TR></TABLE>

Since the last column adds to 452, we have a remainder of 452. 


Since we have a remainder of 452, this means that {{{f(3)=452}}}