Question 117090

Let's simplify this expression using synthetic division



Start with the given expression {{{(5x^3 - 5x + 1)/(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 {{{5x^3}}} to {{{-5x^1}}} there is a zero coefficient for {{{x^2}}}. This is simply because {{{5x^3 - 5x + 1}}} really looks like {{{5x^3+0x^2+-5x^1+1x^0}}}<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>5</TD><TD>0</TD><TD>-5</TD><TD>1</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 5)

<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>5</TD><TD>0</TD><TD>-5</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>5</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 3 by 5 and place the product (which is 15)  right underneath the second  coefficient (which is 0)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>5</TD><TD>0</TD><TD>-5</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>15</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>5</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 15 and 0 to get 15. Place the sum right underneath 15.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>5</TD><TD>0</TD><TD>-5</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>15</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>5</TD><TD>15</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 -5)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>5</TD><TD>0</TD><TD>-5</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>15</TD><TD>45</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>5</TD><TD>15</TD><TD></TD><TD></TD></TR></TABLE>

    Add 45 and -5 to get 40. Place the sum right underneath 45.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>5</TD><TD>0</TD><TD>-5</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>15</TD><TD>45</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>5</TD><TD>15</TD><TD>40</TD><TD></TD></TR></TABLE>

    Multiply 3 by 40 and place the product (which is 120)  right underneath the fourth  coefficient (which is 1)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>5</TD><TD>0</TD><TD>-5</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>15</TD><TD>45</TD><TD>120</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>5</TD><TD>15</TD><TD>40</TD><TD></TD></TR></TABLE>

    Add 120 and 1 to get 121. Place the sum right underneath 120.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>5</TD><TD>0</TD><TD>-5</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>15</TD><TD>45</TD><TD>120</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>5</TD><TD>15</TD><TD>40</TD><TD>121</TD></TR></TABLE>

Since the last column adds to 121, we have a remainder of 121. This means {{{x-3}}} is <b>not</b> a factor of  {{{5x^3 - 5x + 1}}}

Now lets look at the bottom row of coefficients:


The first 3 coefficients (5,15,40) form the quotient


{{{5x^2 + 15x + 40}}}


and the last coefficient 121, is the remainder, which is placed over {{{x-3}}} like this


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




Putting this altogether, we get:


{{{5x^2 + 15x + 40+121/(x-3)}}}


So {{{(5x^3 - 5x + 1)/(x-3)=5x^2 + 15x + 40+121/(x-3)}}}


which looks like this in remainder form:

{{{(5x^3 - 5x + 1)/(x-3)=5x^2 + 15x + 40}}} remainder 121



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