Question 149491
Let's simplify this expression using synthetic division



Start with the given expression {{{(x^3 - 4x^2 + x + 6)/(x+1)}}}


First lets find our test zero:


{{{x+1=0}}} Set the denominator {{{x+1}}} equal to zero


{{{x=-1}}} Solve for x.


so our test zero is -1



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>-1</TD><TD>|</TD><TD>1</TD><TD>-4</TD><TD>1</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>-1</TD><TD>|</TD><TD>1</TD><TD>-4</TD><TD>1</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 -1 by 1 and place the product (which is -1)  right underneath the second  coefficient (which is -4)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-4</TD><TD>1</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</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 -1 and -4 to get -5. Place the sum right underneath -1.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-4</TD><TD>1</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-5</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -1 by -5 and place the product (which is 5)  right underneath the third  coefficient (which is 1)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-4</TD><TD>1</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>5</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-5</TD><TD></TD><TD></TD></TR></TABLE>

    Add 5 and 1 to get 6. Place the sum right underneath 5.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-4</TD><TD>1</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>5</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-5</TD><TD>6</TD><TD></TD></TR></TABLE>

    Multiply -1 by 6 and place the product (which is -6)  right underneath the fourth  coefficient (which is 6)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-4</TD><TD>1</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>5</TD><TD>-6</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-5</TD><TD>6</TD><TD></TD></TR></TABLE>

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

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-4</TD><TD>1</TD><TD>6</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>5</TD><TD>-6</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-5</TD><TD>6</TD><TD>0</TD></TR></TABLE>


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,-5,6) form the quotient


{{{x^2 - 5x + 6}}}



So {{{(x^3 - 4x^2 + x + 6)/(x+1)=x^2 - 5x + 6}}}


Basically  {{{x^3 - 4x^2 + x + 6}}} factors to {{{(x+1)(x^2 - 5x + 6)}}}


Now lets break  {{{x^2 - 5x + 6}}} down further





Looking at the expression {{{x^2-5x+6}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-5}}}, and the last term is {{{6}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{6}}} to get {{{(1)(6)=6}}}.



Now the question is: what two whole numbers multiply to {{{6}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-5}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{6}}} (the previous product).



Factors of {{{6}}}:

1,2,3,6

-1,-2,-3,-6



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{6}}}.

1*6
2*3
(-1)*(-6)
(-2)*(-3)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-5}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>1+6=7</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>2+3=5</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-1+(-6)=-7</font></td></tr><tr><td  align="center"><font color=red>-2</font></td><td  align="center"><font color=red>-3</font></td><td  align="center"><font color=red>-2+(-3)=-5</font></td></tr></table>



From the table, we can see that the two numbers {{{-2}}} and {{{-3}}} add to {{{-5}}} (the middle coefficient).



So the two numbers {{{-2}}} and {{{-3}}} both multiply to {{{6}}} <font size=4><b>and</b></font> add to {{{-5}}}



Now replace the middle term {{{-5x}}} with {{{-2x-3x}}}. Remember, {{{-2}}} and {{{-3}}} add to {{{-5}}}. So this shows us that {{{-2x-3x=-5x}}}.



{{{x^2+highlight(-2x-3x)+6}}} Replace the second term {{{-5x}}} with {{{-2x-3x}}}.



{{{(x^2-2x)+(-3x+6)}}} Group the terms into two pairs.



{{{x(x-2)+(-3x+6)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x-2)-3(x-2)}}} Factor out {{{3}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(x-3)(x-2)}}} Combine like terms. Or factor out the common term {{{x-2}}}




So {{{x^2-5x+6}}} factors to {{{(x-3)(x-2)}}}.



So {{{(x+1)(x^2 - 5x + 6)}}} now becomes {{{(x+1)(x-3)(x-2)}}}


---------------------------------------------



Answer:



So {{{x^3-4x^2+x+6}}} completely factors to {{{(x+1)(x-3)(x-2)}}}