Question 90279
Start with the given polynomial {{{(4x^3 + 5x^2 - 13x - 7)/(x-2)}}}


First lets find our test zero:


{{{x-2=0}}} Set the denominator {{{x-2}}} equal to zero

{{{x=2}}} Solve for x.


so our test zero is 2



Now 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>2</TD><TD>|</TD><TD>4</TD><TD>5</TD><TD>-13</TD><TD>-7</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 (it is the coefficient with the highest exponent which is 4)

<TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>4</TD><TD>5</TD><TD>-13</TD><TD>-7</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 2 by 4 and place the product (which is 8)  right underneath the second  coefficient (which is 5)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>4</TD><TD>5</TD><TD>-13</TD><TD>-7</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>8</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 8 and 5 to get 13. Place the sum right underneath 8.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>4</TD><TD>5</TD><TD>-13</TD><TD>-7</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>8</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>13</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 2 by 13 and place the product (which is 26)  right underneath the third  coefficient (which is -13)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>4</TD><TD>5</TD><TD>-13</TD><TD>-7</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>8</TD><TD>26</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>13</TD><TD></TD><TD></TD></TR></TABLE>

    Add 26 and -13 to get 13. Place the sum right underneath 26.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>4</TD><TD>5</TD><TD>-13</TD><TD>-7</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>8</TD><TD>26</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>13</TD><TD>13</TD><TD></TD></TR></TABLE>

    Multiply 2 by 13 and place the product (which is 26)  right underneath the fourth  coefficient (which is -7)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>4</TD><TD>5</TD><TD>-13</TD><TD>-7</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>8</TD><TD>26</TD><TD>26</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>13</TD><TD>13</TD><TD></TD></TR></TABLE>

    Add 26 and -7 to get 19. Place the sum right underneath 26.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>4</TD><TD>5</TD><TD>-13</TD><TD>-7</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>8</TD><TD>26</TD><TD>26</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>13</TD><TD>13</TD><TD>19</TD></TR></TABLE>

Since the last column adds to 19, we have a remainder of 19. This means {{{x-2}}} is <b>not</b> a factor of  {{{4x^3 + 5x^2 - 13x - 7}}}

Now lets look at the bottom row of coefficients:


The first 3 coefficients (4,13,13) form the quotient


{{{4x^2 + 13x + 13}}}


and the last coefficient 19, is the remainder. 


So the remainder is 19


note: it looks like you added -26 and -7 to get -33. Instead you should add 26 and -7 to get 19






You can use this <a href=http://calc101.com/webMathematica/long-divide.jsp>online polynomial division calculator</a> to check your work