Question 92192


Start with the given polynomial {{{(x^4 + 12x^3 + 6x + 27)/(x-3)}}}


First lets find our test zero:


{{{x-3=0}}} Set the denominator {{{x-3}}} equal to zero


{{{x=3}}} Solve for x.


so 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 numerator to the right of the test zero.(note: remember if a polynomial goes from {{{12x^3}}} to {{{6x^1}}} there is a zero coefficient for {{{x^2}}}. This is simply because {{{x^4 + 12x^3 + 6x + 27}}} really looks like {{{1x^4+12x^3+0x^2+6x^1+27x^0}}}<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</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 1)

<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</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>1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 3 by 1 and place the product (which is 3)  right underneath the second  coefficient (which is 12)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>3</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

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

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>3</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>15</TD><TD></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 0)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>3</TD><TD>45</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>15</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 45 and 0 to get 45. Place the sum right underneath 45.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>3</TD><TD>45</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>15</TD><TD>45</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 3 by 45 and place the product (which is 135)  right underneath the fourth  coefficient (which is 6)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>3</TD><TD>45</TD><TD>135</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>15</TD><TD>45</TD><TD></TD><TD></TD></TR></TABLE>

    Add 135 and 6 to get 141. Place the sum right underneath 135.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>3</TD><TD>45</TD><TD>135</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>15</TD><TD>45</TD><TD>141</TD><TD></TD></TR></TABLE>

    Multiply 3 by 141 and place the product (which is 423)  right underneath the fifth  coefficient (which is 27)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>3</TD><TD>45</TD><TD>135</TD><TD>423</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>15</TD><TD>45</TD><TD>141</TD><TD></TD></TR></TABLE>

    Add 423 and 27 to get 450. Place the sum right underneath 423.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>1</TD><TD>12</TD><TD>0</TD><TD>6</TD><TD>27</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>3</TD><TD>45</TD><TD>135</TD><TD>423</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>15</TD><TD>45</TD><TD>141</TD><TD>450</TD></TR></TABLE>

Since the last column adds to 450, we have a remainder of 450. This means {{{x-3}}} is <b>not</b> a factor of  {{{x^4 + 12x^3 + 6x + 27}}}

Now lets look at the bottom row of coefficients:


The first 4 coefficients (1,15,45,141) form the quotient


{{{x^3 + 15x^2 + 45x + 141}}}


and the last coefficient 450, is the remainder, which is placed over {{{x-3}}} like this


{{{450/(x-3)}}}




Putting this altogether, we get:


{{{x^3 + 15x^2 + 45x + 141+450/(x-3)}}}


So {{{(x^4 + 12x^3 + 6x + 27)/(x-3)=x^3 + 15x^2 + 45x + 141+450/(x-3)}}}


which looks like this in remainder form:

{{{(x^4 + 12x^3 + 6x + 27)/(x-3)=x^3 + 15x^2 + 45x + 141}}} remainder 450



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