<PRE><B>Question 123010</B>


Any rational zero can be found through this equation


*[Tex \LARGE Roots=\frac{p}{q}] where p and q are the factors of the last and first coefficients



So let&#39;s list the factors of -40 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm2, \pm4, \pm5, \pm8, \pm10, \pm20, \pm40]


Now let&#39;s list the factors of 1 (the first coefficient):


*[Tex \LARGE q=\pm1]


Now let&#39;s divide each factor of the last coefficient by each factor of the first coefficient



*[Tex \LARGE \frac{1}{1}, \frac{2}{1}, \frac{4}{1}, \frac{5}{1}, \frac{8}{1}, \frac{10}{1}, \frac{20}{1}, \frac{40}{1}, \frac{-1}{1}, \frac{-2}{1}, \frac{-4}{1}, \frac{-5}{1}, \frac{-8}{1}, \frac{-10}{1}, \frac{-20}{1}, \frac{-40}{1}]







Now simplify


These are all the distinct rational zeros of the function that could occur


*[Tex \LARGE  1, 2, 4, 5, 8, 10, 20, 40, -1, -2, -4, -5, -8, -10, -20, -40]



To save time, I&#39;m only going to use synthetic division on the possible zeros that are actually zeros of the function.
Otherwise, I would have to use synthetic division on every possible root (there are 16 possible roots, so that means there would be at most 16 synthetic division tables).
However, you might be required to follow this procedure, so this is why I&#39;m showing you how to set up a problem like this




When you graph this polynomial, you will see that {{{x=2}}} is a zero. So we&#39;ll use this for the synthetic division  





Now set up the synthetic division table by placing the zero in the upper left corner and placing the coefficients of the polynomial to the right of the test zero.&lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;7&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;-40&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)

&lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;7&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;-40&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Multiply 2 by 1 and place the product (which is 2)  right underneath the second  coefficient (which is 7)

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;7&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;-40&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Add 2 and 7 to get 9. Place the sum right underneath 2.

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;7&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;-40&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;9&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Multiply 2 by 9 and place the product (which is 18)  right underneath the third  coefficient (which is 2)

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;7&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;-40&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;18&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;9&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Add 18 and 2 to get 20. Place the sum right underneath 18.

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;7&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;-40&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;18&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;9&lt;/TD&gt;&lt;TD&gt;20&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Multiply 2 by 20 and place the product (which is 40)  right underneath the fourth  coefficient (which is -40)

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;7&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;-40&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;18&lt;/TD&gt;&lt;TD&gt;40&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;9&lt;/TD&gt;&lt;TD&gt;20&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Add 40 and -40 to get 0. Place the sum right underneath 40.

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;7&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;-40&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;18&lt;/TD&gt;&lt;TD&gt;40&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;9&lt;/TD&gt;&lt;TD&gt;20&lt;/TD&gt;&lt;TD&gt;0&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

Since the last column adds to zero, we have a remainder of zero. This means {{{x-2}}} is a factor of  {{{x^3 + 7x^2 + 2x - 40}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,9,20) form the quotient


{{{x^2 + 9x + 20}}}



So {{{(x^3 + 7x^2 + 2x - 40)/(x-2)=x^2 + 9x + 20}}}


You can use this &lt;a href=http://calc101.com/webMathematica/long-divide.jsp&gt;online polynomial division calculator&lt;/a&gt; to check your work


Basically  {{{x^3 + 7x^2 + 2x - 40}}} factors to {{{(x-2)(x^2 + 9x + 20)}}}


Now lets break  {{{x^2 + 9x + 20}}} down further






Looking at {{{x^2+9x+20}}} we can see that the first term is {{{x^2}}} and the last term is {{{20}}} where the coefficients are 1 and 20 respectively.


Now multiply the first coefficient 1 and the last coefficient 20 to get 20. Now what two numbers multiply to 20 and add to the  middle coefficient 9? Let&#39;s list all of the factors of 20:




Factors of 20:

1,2,4,5,10,20


-1,-2,-4,-5,-10,-20 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 20

1*20

2*10

4*5

(-1)*(-20)

(-2)*(-10)

(-4)*(-5)


note: remember two negative numbers multiplied together make a positive number



Now which of these pairs add to 9? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 9


&lt;table border=&quot;1&quot;&gt;&lt;th&gt;First Number&lt;/th&gt;&lt;th&gt;Second Number&lt;/th&gt;&lt;th&gt;Sum&lt;/th&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;1&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;20&lt;/td&gt;&lt;td&gt;1+20=21&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;2&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;10&lt;/td&gt;&lt;td&gt;2+10=12&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;4&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;5&lt;/td&gt;&lt;td&gt;4+5=9&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;-1&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;-20&lt;/td&gt;&lt;td&gt;-1+(-20)=-21&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;-2&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;-10&lt;/td&gt;&lt;td&gt;-2+(-10)=-12&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;-4&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;-5&lt;/td&gt;&lt;td&gt;-4+(-5)=-9&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From this list we can see that 4 and 5 add up to 9 and multiply to 20



Now looking at the expression {{{1x^2+9x+20}}}, replace {{{9x}}} with {{{4x+5x}}} (notice {{{4x+5x}}} adds up to {{{9x}}}. So it is equivalent to {{{9x}}})


{{{x^2+highlight(4x+5x)+20}}}



Now let&#39;s factor {{{1x^2+4x+5x+20}}} by grouping:



{{{(x^2+4x)+(5x+20)}}} Group like terms



{{{x(x+4)+5(x+4)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



{{{(x+5)(x+4)}}} Since we have a common term of {{{x+4}}}, we can combine like terms



So {{{x^2+9x+20}}} factors to {{{(x+5)(x+4)}}}



{{{(x-2)(x+5)(x+4)}}} Now reintroduce the factor {{{x-2}}}






Now set each factor equal to zero:


{{{x-2=0}}}, {{{x+5=0}}} or {{{x+4=0}}}


Now solve for x for each factor:


{{{x=2}}}, {{{x=-5}}} or {{{x=-4}}}






------------------------------------------------------------




     Answer:




So the zeros of {{{x^3+7x^2+2x-40}}} are {{{x=2}}}, {{{x=-5}}} or {{{x=-4}}}

</PRE>