Question 1198584
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First let's pull out the GCF x
x^5-13x^3+36x = 0
x(x^4-13x^2+36) = 0


Now let's factor x^4-13x^2+36
To do so, we have at least two methods


Method 1) 
Let w = x^2
So w^2 = x^4


Then x^4-13x^2+36 is the same as w^2-13w+36
Through trial and error, that would factor to (w-4)(w-9)
Note: -4 and -9 add to -13 and multiply to 36.


Then,
(w-4)(w-9) = (x^2-4)(x^2-9)
(w-4)(w-9) = (x-2)(x+2)(x-3)(x+3)
after applying the difference of squares rule.


So overall,
x^5-13x^3+36x = 0
x(x^4-13x^2+36) = 0
x(x-2)(x+2)(x-3)(x+3) = 0
To find the roots, set each factor equal to zero and solve for x.
Example: x-2 = 0 leads to x = 2 as one root.


Therefore, the five roots are: 
{{{x = 0}}}, {{{x = "" +- 2}}}, {{{x = "" +- 3}}}
which is the shorthand way of saying
x = 0, x = -2, x = 2, x = -3, x = 3


Answer: Choice A


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Method 2)


Here's another way to factor x^4-13x^2+36


Use the rational root theorem to list out the factors of the last term 36: 
1, 2, 3, 4, 6, 9, 12, 18, 36
List the negative factors as well
-1, -2, -3, -4, -6, -9, -12, -18, -36


Then through trial and error, you should find that  x = -2, x = 2, x = -3, x = 3 are actual roots since they cause x^4-13x^2+36 to be zero.
This gives the factorization (x-2)(x+2)(x-3)(x+3)


I don't recommend this second method as the first method is more efficient. But it's good to keep an alternative in mind.
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