Question 189561

Let's simplify this expression using synthetic division



Start with the given expression {{{(2x^3 + 4x^2 - 3x + 10)/(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.<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>4</TD><TD>-3</TD><TD>10</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 2)

<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>4</TD><TD>-3</TD><TD>10</TD></TR><TR><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></TR></TABLE>

    Multiply 3 by 2 and place the product (which is 6)  right underneath the second  coefficient (which is 4)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>4</TD><TD>-3</TD><TD>10</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>2</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 6 and 4 to get 10. Place the sum right underneath 6.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>4</TD><TD>-3</TD><TD>10</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>2</TD><TD>10</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 3 by 10 and place the product (which is 30)  right underneath the third  coefficient (which is -3)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>4</TD><TD>-3</TD><TD>10</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>30</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>10</TD><TD></TD><TD></TD></TR></TABLE>

    Add 30 and -3 to get 27. Place the sum right underneath 30.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>4</TD><TD>-3</TD><TD>10</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>30</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>10</TD><TD>27</TD><TD></TD></TR></TABLE>

    Multiply 3 by 27 and place the product (which is 81)  right underneath the fourth  coefficient (which is 10)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>4</TD><TD>-3</TD><TD>10</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>30</TD><TD>81</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>10</TD><TD>27</TD><TD></TD></TR></TABLE>

    Add 81 and 10 to get 91. Place the sum right underneath 81.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>4</TD><TD>-3</TD><TD>10</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>30</TD><TD>81</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>10</TD><TD>27</TD><TD>91</TD></TR></TABLE>

Since the last column adds to 91, we have a remainder of 91. This means {{{x-3}}} is <b>not</b> a factor of  {{{2x^3 + 4x^2 - 3x + 10}}}

Now lets look at the bottom row of coefficients:


The first 3 coefficients (2,10,27) form the quotient


{{{2x^2 + 10x + 27}}}


and the last coefficient 91, is the remainder, which is placed over {{{x-3}}} like this


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




Putting this altogether, we get:



{{{2x^2 + 10x + 27+91/(x-3)}}}



So {{{(2x^3 + 4x^2 - 3x + 10)/(x-3)=2x^2 + 10x + 27+91/(x-3)}}}



which looks like this in remainder form:



{{{(2x^3 + 4x^2 - 3x + 10)/(x-3)=2x^2 + 10x + 27}}} remainder 91



So the answer is C