Question 117393


Let's set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of {{{x^3 + 6x^2 + 11x + 6}}} to the right of the test zero.<TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>11</TD><TD>6</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>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>11</TD><TD>6</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 -2 by 1 and place the product (which is -2)  right underneath the second  coefficient (which is 6)

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>11</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</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 -2 and 6 to get 4. Place the sum right underneath -2.

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>11</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -2 by 4 and place the product (which is -8)  right underneath the third  coefficient (which is 11)

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>11</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>-8</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD></TD><TD></TD></TR></TABLE>

    Add -8 and 11 to get 3. Place the sum right underneath -8.

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>11</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>-8</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD>3</TD><TD></TD></TR></TABLE>

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

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>11</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>-8</TD><TD>-6</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD>3</TD><TD></TD></TR></TABLE>

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

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>11</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>-8</TD><TD>-6</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD>3</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x+2}}} is a factor of  {{{x^3 + 6x^2 + 11x + 6}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,4,3) form the quotient


{{{x^2 + 4x + 3}}}



So {{{(x^3 + 6x^2 + 11x + 6)/(x+2)=x^2 + 4x + 3}}}


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


Basically  {{{x^3 + 6x^2 + 11x + 6}}} factors to {{{(x+2)(x^2 + 4x + 3)}}}


Now lets break  {{{x^2 + 4x + 3}}} down further






Looking at {{{x^2+4x+3}}} we can see that the first term is {{{x^2}}} and the last term is {{{3}}} where the coefficients are 1 and 3 respectively.


Now multiply the first coefficient 1 and the last coefficient 3 to get 3. Now what two numbers multiply to 3 and add to the  middle coefficient 4? Let's list all of the factors of 3:




Factors of 3:

1,3


-1,-3 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 3

1*3

(-1)*(-3)


note: remember two negative numbers multiplied together make a positive number



Now which of these pairs add to 4? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 4


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">3</td><td>1+3=4</td></tr><tr><td align="center">-1</td><td align="center">-3</td><td>-1+(-3)=-4</td></tr></table>



From this list we can see that 1 and 3 add up to 4 and multiply to 3



Now looking at the expression {{{x^2+4x+3}}}, replace {{{4x}}} with {{{1x+3x}}} (notice {{{1x+3x}}} adds up to {{{4x}}}. So it is equivalent to {{{4x}}})


{{{x^2+highlight(1x+3x)+3}}}



Now let's factor {{{x^2+1x+3x+3}}} by grouping:



{{{(x^2+1x)+(3x+3)}}} Group like terms



{{{x(x+1)+3(x+1)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{3}}} out of the second group



{{{(x+3)(x+1)}}} Since we have a common term of {{{x+1}}}, we can combine like terms


So {{{x^2+1x+3x+3}}} factors to {{{(x+3)(x+1)}}}



So this also means that {{{x^2+4x+3}}} factors to {{{(x+3)(x+1)}}} (since {{{x^2+4x+3}}} is equivalent to {{{x^2+1x+3x+3}}})


So {{{x^2+4x+3}}} factors to {{{(x+3)(x+1)}}}



{{{(x+2)(x+3)(x+1)}}} Now reintroduce the first factor {{{x+2}}}


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Answer:


So {{{x^3 + 6x^2 + 11x + 6}}} factors to {{{(x+2)(x+3)(x+1)}}} which means the answer is D)


Notice if we graph {{{x^3 + 6x^2 + 11x + 6}}} we get


{{{ graph( 500, 500, -10, 10, -10, 10, x^3 + 6x^2 + 11x + 6) }}} Graph of {{{x^3 + 6x^2 + 11x + 6}}}



and if we graph {{{(x+2)(x+3)(x+1)}}}, we get



{{{ graph( 500, 500, -10, 10, -10, 10, (x+2)(x+3)(x+1)) }}} Graph of {{{(x+2)(x+3)(x+1)}}}


and you can see that the two graphs would overlap each other if they were plotted on the same screen. So this means that the two polynomials are equivalent. So this visually verifies our answer.