Question 85753


Start with the given polynomial {{{(2x^4 - 10x^2 + 6x - 8)/(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 upperleft corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from {{{2x^4}}} to {{{-10x^2}}} there is a zero coefficient. This is simply because {{{2x^4 - 10x^2 + 6x - 8}}} really looks like {{{2x^4+0x^3+-10x^2+6x^1+-8x^0}}}<TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</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 2)

<TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</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>2</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -3 by 2 and place the product (which is -6)  right underneath 0

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</TD></TR><TR><TD></TD><TD></TD><TD></TD><TD>-6</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -6 and 0 to get -6. Place the sum right underneath -6.

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</TD></TR><TR><TD></TD><TD></TD><TD></TD><TD>-6</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-6</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -3 by -6 and place the product (which is 18)  right underneath -10

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</TD></TR><TR><TD></TD><TD></TD><TD></TD><TD>-6</TD><TD>18</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-6</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 18 and -10 to get 8. Place the sum right underneath 18.

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</TD></TR><TR><TD></TD><TD></TD><TD></TD><TD>-6</TD><TD>18</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-6</TD><TD>8</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -3 by 8 and place the product (which is -24)  right underneath 6

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</TD></TR><TR><TD></TD><TD></TD><TD></TD><TD>-6</TD><TD>18</TD><TD>-24</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-6</TD><TD>8</TD><TD></TD><TD></TD></TR></TABLE>

    Add -24 and 6 to get -18. Place the sum right underneath -24.

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</TD></TR><TR><TD></TD><TD></TD><TD></TD><TD>-6</TD><TD>18</TD><TD>-24</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-6</TD><TD>8</TD><TD>-18</TD><TD></TD></TR></TABLE>

    Multiply -3 by -18 and place the product (which is 54)  right underneath -8

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</TD></TR><TR><TD></TD><TD></TD><TD></TD><TD>-6</TD><TD>18</TD><TD>-24</TD><TD>54</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-6</TD><TD>8</TD><TD>-18</TD><TD></TD></TR></TABLE>

    Add 54 and -8 to get 46. Place the sum right underneath 54.

    <TABLE cellpadding=10><TR><TD>-3</TD><TD>|</TD><TD>2</TD><TD>0</TD><TD>-10</TD><TD>6</TD><TD>-8</TD></TR><TR><TD></TD><TD></TD><TD></TD><TD>-6</TD><TD>18</TD><TD>-24</TD><TD>54</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-6</TD><TD>8</TD><TD>-18</TD><TD>46</TD></TR></TABLE>

Since the last column adds to 46, we have a remainder of 46. This means {{{x+3}}} is a <b>not</b> factor of  {{{2x^4 - 10x^2 + 6x - 8}}}

Now lets look at the bottom row of coefficients:


The first 4 coefficients (2,-6,8,-18) form the quotient


{{{2x^3 - 6x^2 + 8x - 18}}}


and the last coefficient 46, is the remainder, which is placed over {{{x+3}}} like this


{{{46/(x+3)}}}

Putting this altogether, we get:


{{{2x^3 - 6x^2 + 8x - 18+46/(x+3)}}}


So {{{(2x^4 - 10x^2 + 6x - 8)/(x+3)=2x^3 - 6x^2 + 8x - 18+46/(x+3)}}}


which looks like this in remainder form:

{{{(2x^4 - 10x^2 + 6x - 8)/(x+3)=2x^3 - 6x^2 + 8x - 18}}} remainder 46