Question 124378
Let's simplify this expression using synthetic division



Start with the given expression {{{(2x^4 - 5x^3 + 7x^2 - 3x + 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.<TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>7</TD><TD>-3</TD><TD>1</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>-5</TD><TD>7</TD><TD>-3</TD><TD>1</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 the second  coefficient (which is -5)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>7</TD><TD>-3</TD><TD>1</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 -5 to get 1. Place the sum right underneath 6.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>7</TD><TD>-3</TD><TD>1</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>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 3 by 1 and place the product (which is 3)  right underneath the third  coefficient (which is 7)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>7</TD><TD>-3</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>3</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 3 and 7 to get 10. Place the sum right underneath 3.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>7</TD><TD>-3</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>3</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD>10</TD><TD></TD><TD></TD></TR></TABLE>

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

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>7</TD><TD>-3</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>3</TD><TD>30</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</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>-5</TD><TD>7</TD><TD>-3</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>3</TD><TD>30</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</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 fifth  coefficient (which is 1)

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>7</TD><TD>-3</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>3</TD><TD>30</TD><TD>81</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD>10</TD><TD>27</TD><TD></TD></TR></TABLE>

    Add 81 and 1 to get 82. Place the sum right underneath 81.

    <TABLE cellpadding=10><TR><TD>3</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>7</TD><TD>-3</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>6</TD><TD>3</TD><TD>30</TD><TD>81</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD>10</TD><TD>27</TD><TD>82</TD></TR></TABLE>

Since the last column adds to 82, we have a remainder of 82. This means {{{x-3}}} is <b>not</b> a factor of  {{{2x^4 - 5x^3 + 7x^2 - 3x + 1}}}

Now lets look at the bottom row of coefficients:


Looking at the last line, we see the coefficients: 2,1,10, and 27



So the coefficients for the quotient are <b>not</b> 2,1,8, and 21



So the answer is false and you are correct.