Question 86302
Let P be the unknown polynomial

Lets assume that {{{x-4}}} is a factor


{{{x^3-37x+84=(x-4)(P)}}} Start with the given equation


Divide both sides by {{{x-4}}}

{{{(x^3-37x+84)/(x-4)=P}}}


Now lets divide {{{(x^3 - 37x + 84)/(x-4)}}}


First lets find our test zero:


{{{x-4=0}}} Set the denominator {{{x-4}}} equal to zero

{{{x=4}}} Solve for x.


so our test zero is 4



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 {{{1x^3}}} to {{{-37x^1}}} there is a zero coefficient for {{{x^2}}}. This is simply because {{{x^3 - 37x + 84}}} really looks like {{{1x^3+0x^2+-37x^1+84x^0}}}<TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-37</TD><TD>84</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>4</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-37</TD><TD>84</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 4 by 1 and place the product (which is 4)  right underneath the second  coefficient (which is 0)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-37</TD><TD>84</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</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 4 and 0 to get 4. Place the sum right underneath 4.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-37</TD><TD>84</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</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 4 by 4 and place the product (which is 16)  right underneath the third  coefficient (which is -37)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-37</TD><TD>84</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>16</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD></TD><TD></TD></TR></TABLE>

    Add 16 and -37 to get -21. Place the sum right underneath 16.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-37</TD><TD>84</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>16</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD>-21</TD><TD></TD></TR></TABLE>

    Multiply 4 by -21 and place the product (which is -84)  right underneath the fourth  coefficient (which is 84)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-37</TD><TD>84</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>16</TD><TD>-84</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD>-21</TD><TD></TD></TR></TABLE>

    Add -84 and 84 to get 0. Place the sum right underneath -84.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-37</TD><TD>84</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>16</TD><TD>-84</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD>-21</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x-4}}} is a factor of  {{{x^3 - 37x + 84}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,4,-21) form the quotient


{{{x^2 + 4x - 21}}}


So {{{(x^3 - 37x + 84)/(x-4)=x^2 + 4x - 21}}}


So the unknown polynomial P is 


{{{P=x^2 + 4x - 21}}}


Now multiply both sides by {{{x-4}}}


{{{cross(x-4)(x^3 - 37x + 84)/cross(x-4)=(x^2 + 4x - 21)(x-4)}}}


So the expression factors to this


{{{x^3 - 37x + 84=(x^2 + 4x - 21)(x-4)}}}