Question 191251
First, perform synthetic division where -2 is the test zero (let me know if you need help with synthetic division)


<pre>
-2 | 1   4   0   -16   -16
   |    -2  -4     8    16
   ------------------------
     1   2  -4    -8     0
</pre>



Since the last number in the bottom row is zero, this means that the remainder is 0. So -2 is a root of {{{P(x)= x^4+4x^3-16x-16}}}


The first 4 numbers form the depressed polynomial {{{x^3+2x^2-4x-8}}}. This means that {{{x^4+4x^3-16x-16=(x+2)(x^3+2x^2-4x-8)}}}

=================================



Now perform synthetic division on {{{x^3+2x^2-4x-8}}} using the same test zero:


<pre>
-2 | 1   2  -4    -8   
   |    -2   0     8   
   ------------------
     1   0  -4     0   
</pre>


Notice how the last number in the bottom row is 0. So -2 is a root of {{{x^3+2x^2-4x-8}}}. So far, r=-2 is a root of multiplicity 2 (ie -2 is a root twice).



The first 3 numbers in the bottom row form the new polynomial {{{x^2-4}}}. This tells us that {{{x^3+2x^2-4x-8=(x+2)(x^2-4)}}}



So {{{x^4+4x^3-16x-16=(x+2)(x^3+2x^2-4x-8)=(x+2)(x+2)(x^2-4)}}}


=================================



Now perform synthetic division on the polynomial {{{x^2-4}}}

<pre>
-2 | 1   0  -4     
   |    -2   4    
   ------------
     1  -2   0

</pre>



So -2 is a root of {{{x^2-4}}}. So this means that r=-2 is a root of multiplicity 3 (ie -2 is a root three times).



The first two numbers in the bottom row form the new polynomial: {{{x-2}}}



Now because -2 is NOT a root of {{{x-2}}}, this means that we can stop looking for more roots of -2.



So {{{x^4+4x^3-16x-16=(x+2)(x^3+2x^2-4x-8)=(x+2)(x+2)(x^2-4)=(x+2)(x+2)(x+2)(x-2)}}}


Or in other words, {{{x^4+4x^3-16x-16=(x+2)(x+2)(x+2)(x-2)}}}



Notice how the factor {{{x+2}}} is repeated 3 times, this supports our conclusion that r=-2 is a root of multiplicity 3