Question 133085

Let's simplify this expression using synthetic division



Start with the given expression {{{(y^3 + 8y^2 + 14y + 12)/(y+6)}}}


First lets find our test zero:


{{{y+6=0}}} Set the denominator {{{y+6}}} equal to zero


{{{y=-6}}} Solve for y.


so our test zero is -6



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>-6</TD><TD>|</TD><TD>1</TD><TD>8</TD><TD>14</TD><TD>12</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 1)

<TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>8</TD><TD>14</TD><TD>12</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 -6 by 1 and place the product (which is -6)  right underneath the second  coefficient (which is 8)

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>8</TD><TD>14</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</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 -6 and 8 to get 2. Place the sum right underneath -6.

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>8</TD><TD>14</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>2</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -6 by 2 and place the product (which is -12)  right underneath the third  coefficient (which is 14)

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>8</TD><TD>14</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>-12</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>2</TD><TD></TD><TD></TD></TR></TABLE>

    Add -12 and 14 to get 2. Place the sum right underneath -12.

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>8</TD><TD>14</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>-12</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>2</TD><TD>2</TD><TD></TD></TR></TABLE>

    Multiply -6 by 2 and place the product (which is -12)  right underneath the fourth  coefficient (which is 12)

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>8</TD><TD>14</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>-12</TD><TD>-12</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>2</TD><TD>2</TD><TD></TD></TR></TABLE>

    Add -12 and 12 to get 0. Place the sum right underneath -12.

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>8</TD><TD>14</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>-12</TD><TD>-12</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>2</TD><TD>2</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{y+6}}} is a factor of  {{{y^3 + 8y^2 + 14y + 12}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,2,2) form the quotient


{{{y^2 + 2y + 2}}}



So {{{(y^3 + 8y^2 + 14y + 12)/(y+6)=y^2 + 2y + 2}}}


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