Question 90671
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's list the factors of 15:


*[Tex \LARGE p=\pm1, \pm3, \pm5, \pm15]


Now let's list the factors of 1:


*[Tex \LARGE q=\pm1]


Now let's divide each factor of the last coefficient by each factor of the first coefficient



*[Tex \LARGE \frac{1}{1}, \frac{3}{1}, \frac{5}{1}, \frac{15}{1}, \frac{-1}{1}, \frac{-3}{1}, \frac{-5}{1}, \frac{-15}{1}]


Now simplify


These are all the possible zeros of the function


*[Tex \LARGE 1, 3, 5, 15, -1, -3, -5, -15]



To save time, I'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 8 possible roots, so that means there would be at most 8 synthetic division tables).
However, you might be required to follow this procedure, so this is why I'm showing you how to set up a problem like this



If you're not required to follow this procedure, simply use a graphing calculator to find the roots



So with a graphing calculator, we find one root at x=-1



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.(note: remember if a polynomial goes from {{{1x^4}}} to {{{-16x^2}}} there is a zero coefficient for {{{x^3}}}. This is simply because {{{x^4 - 16x^2 + 15}}} really looks like {{{1x^4+0x^3+-16x^2+0x^1+15x^0}}}<TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</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 1)

<TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</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>1</TD><TD></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 0)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -1 and 0 to get -1. Place the sum right underneath -1.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</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 third  coefficient (which is -16)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 1 and -16 to get -15. Place the sum right underneath 1.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -1 by -15 and place the product (which is 15)  right underneath the fourth  coefficient (which is 0)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>1</TD><TD>15</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD></TD><TD></TD></TR></TABLE>

    Add 15 and 0 to get 15. Place the sum right underneath 15.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>1</TD><TD>15</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD>15</TD><TD></TD></TR></TABLE>

    Multiply -1 by 15 and place the product (which is -15)  right underneath the fifth  coefficient (which is 15)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>1</TD><TD>15</TD><TD>-15</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD>15</TD><TD></TD></TR></TABLE>

    Add -15 and 15 to get 0. Place the sum right underneath -15.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-16</TD><TD>0</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>1</TD><TD>15</TD><TD>-15</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD>15</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x+1}}} is a factor of  {{{x^4 - 16x^2 + 15}}}


Now lets look at the bottom row of coefficients:


The first 4 coefficients (1,-1,-15,15) form the quotient


{{{x^3 - x^2 - 15x + 15}}}



So {{{(x^4 - 16x^2 + 15)/(x+1)=x^3 - x^2 - 15x + 15}}}


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




Using the same technique above, we find another zero at x=1

Now lets perform synthetic division on {{{x^3 - x^2 - 15x + 15}}} with the test zero of x=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>-1</TD><TD>-15</TD><TD>15</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>-1</TD><TD>-15</TD><TD>15</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 -1)

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD>15</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 -1 to get 0. Place the sum right underneath 1.

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD>15</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>0</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 1 by 0 and place the product (which is 0)  right underneath the third  coefficient (which is -15)

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>1</TD><TD>0</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD></TD><TD></TD></TR></TABLE>

    Add 0 and -15 to get -15. Place the sum right underneath 0.

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>1</TD><TD>0</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD>-15</TD><TD></TD></TR></TABLE>

    Multiply 1 by -15 and place the product (which is -15)  right underneath the fourth  coefficient (which is 15)

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>1</TD><TD>0</TD><TD>-15</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD>-15</TD><TD></TD></TR></TABLE>

    Add -15 and 15 to get 0. Place the sum right underneath -15.

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>1</TD><TD>-1</TD><TD>-15</TD><TD>15</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>1</TD><TD>0</TD><TD>-15</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD>-15</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x-1}}} is a factor of  {{{x^3 - x^2 - 15x + 15}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,0,-15) form the quotient


{{{x^2 - 15}}}



So {{{(x^3 - x^2 - 15x + 15)/(x-1)=x^2 - 15}}}


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



Now set {{{x^2 - 15}}} equal to zero


{{{x^2 - 15=0}}}


{{{x^2 = 15}}}


{{{x = 0+-sqrt(15)}}} Take the square root of both sides


So two roots are

{{{x =sqrt(15)}}} or {{{x = -sqrt(15)}}}



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Answer:


So the roots are {{{x=-1}}}, {{{x=1}}}, {{{x =sqrt(15)}}} or {{{x = -sqrt(15)}}}