Question 189557

First set up the synthetic division table


<TABLE cellpadding=10><TR><TD>i</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>0</TD><TD>-4</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>


Start by bringing down the leading coefficient 1


<TABLE cellpadding=10><TR><TD>i</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>0</TD><TD>-4</TD></TR>
<TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR>
<TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>


    Multiply i by 1 and place the product (which is i)  right underneath the second  coefficient (which is 2)

<TABLE cellpadding=10><TR><TD>i</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>0</TD><TD>-4</TD></TR>
<TR><TD></TD><TD>|</TD><TD></TD><TD>i</TD><TD></TD><TD></TD><TD></TD></TR>
<TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add i and 2 to get i+2. Place the sum right underneath i.

<TABLE cellpadding=10><TR><TD>i</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>0</TD><TD>-4</TD></TR>
<TR><TD></TD><TD>|</TD><TD></TD><TD>i</TD><TD></TD><TD></TD><TD></TD></TR>
<TR><TD></TD><TD></TD><TD>1</TD><TD>i+2</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply i by i+2 to get {{{i^2+2i=-1+2i}}} and place it right underneath the third  coefficient 0

<TABLE cellpadding=10><TR><TD>i</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>0</TD><TD>-4</TD></TR>
<TR><TD></TD><TD>|</TD><TD></TD><TD>i</TD><TD>-1+2i</TD><TD></TD><TD></TD></TR>
<TR><TD></TD><TD></TD><TD>1</TD><TD>i+2</TD><TD></TD><TD></TD></TR></TABLE>

    Add -1+2i and 0 to get -1+2i. Place the sum right underneath -1+2i.

<TABLE cellpadding=10><TR><TD>i</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>0</TD><TD>-4</TD></TR>
<TR><TD></TD><TD>|</TD><TD></TD><TD>i</TD><TD>-1+2i</TD><TD></TD><TD></TD></TR>
<TR><TD></TD><TD></TD><TD>1</TD><TD>i+2</TD><TD>-1+2i</TD><TD></TD></TR></TABLE>

    Multiply i by -1+2i to get {{{-i+2i^2=-2-i}}} and place the product right underneath the fourth  coefficient -4

<TABLE cellpadding=10><TR><TD>i</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>0</TD><TD>-4</TD></TR>
<TR><TD></TD><TD>|</TD><TD></TD><TD>i</TD><TD>-1+2i</TD><TD>-2-i</TD><TD></TD></TR>
<TR><TD></TD><TD></TD><TD>1</TD><TD>i+2</TD><TD>-1+2i</TD><TD></TD></TR></TABLE>

    Add -2-i and -4 to get -6-i. Place the sum right underneath -2-i.

<TABLE cellpadding=10><TR><TD>i</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>0</TD><TD>-4</TD></TR>
<TR><TD></TD><TD>|</TD><TD></TD><TD>i</TD><TD>-1+2i</TD><TD>-2-i</TD><TD></TD></TR>
<TR><TD></TD><TD></TD><TD>1</TD><TD>i+2</TD><TD>-1+2i</TD><TD>-6-i</TD></TR></TABLE>




Since the remainder is -6-i, this means that {{{f(i)=-6-i}}}



In other words, if we plug "i" in for "x" into {{{f(x)=x^3+2x^2-4}}}, we'll get {{{-6-i}}}