Question 166556

Let's simplify this expression using synthetic division



Start with the given expression {{{(4x^3 - 6x + 5)/(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 {{{4x^3}}} to {{{-6x^1}}} there is a zero coefficient for {{{x^2}}}. This is simply because {{{4x^3 - 6x + 5}}} really looks like {{{4x^3+0x^2+-6x^1+5x^0}}}<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>4</TD><TD>0</TD><TD>-6</TD><TD>5</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>3</TD><TD>|</TD><TD>4</TD><TD>0</TD><TD>-6</TD><TD>5</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 3 by 4 and place the product (which is 12)  right underneath the second  coefficient (which is 0)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>4</TD><TD>0</TD><TD>-6</TD><TD>5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>12</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 12 and 0 to get 12. Place the sum right underneath 12.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>4</TD><TD>0</TD><TD>-6</TD><TD>5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>12</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>12</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 3 by 12 and place the product (which is 36)  right underneath the third  coefficient (which is -6)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>4</TD><TD>0</TD><TD>-6</TD><TD>5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>12</TD><TD>36</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>12</TD><TD></TD><TD></TD></TR></TABLE>

    Add 36 and -6 to get 30. Place the sum right underneath 36.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>4</TD><TD>0</TD><TD>-6</TD><TD>5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>12</TD><TD>36</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>12</TD><TD>30</TD><TD></TD></TR></TABLE>

    Multiply 3 by 30 and place the product (which is 90)  right underneath the fourth  coefficient (which is 5)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>4</TD><TD>0</TD><TD>-6</TD><TD>5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>12</TD><TD>36</TD><TD>90</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>12</TD><TD>30</TD><TD></TD></TR></TABLE>

    Add 90 and 5 to get 95. Place the sum right underneath 90.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>4</TD><TD>0</TD><TD>-6</TD><TD>5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>12</TD><TD>36</TD><TD>90</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>4</TD><TD>12</TD><TD>30</TD><TD>95</TD></TR></TABLE>

Since the last column adds to 95, we have a remainder of 95. This means {{{x-3}}} is <b>not</b> a factor of  {{{4x^3 - 6x + 5}}}

Now lets look at the bottom row of coefficients:


The first 3 coefficients (4,12,30) form the quotient


{{{4x^2 + 12x + 30}}}


and the last coefficient 95, is the remainder, which is placed over {{{x-3}}} like this


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




Putting this altogether, we get:


{{{4x^2 + 12x + 30+95/(x-3)}}}


So {{{(4x^3 - 6x + 5)/(x-3)=4x^2 + 12x + 30+95/(x-3)}}}


which looks like this in remainder form:

{{{(4x^3 - 6x + 5)/(x-3)=4x^2 + 12x + 30}}} remainder 95