SOLUTION: Solve the following polynomial equation x^5-13x^3+36x=0 One of the following is the correct answer. Which one? A) x=0, x=+or-2, x=+or-3 B) x=1, x=2, x=18 C) x=+or-2, x

Algebra.Com
Question 1198584: Solve the following polynomial equation x^5-13x^3+36x=0
One of the following is the correct answer. Which one?
A) x=0, x=+or-2, x=+or-3
B) x=1, x=2, x=18
C) x=+or-2, x+or-3
D) x=+or-3, x=+or-5
E) x=0, x=+or-3, x=+or-12

Found 3 solutions by MathLover1, MathTherapy, math_tutor2020:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

..........factor


factor :
let

Substitute back

so,

solutions:






answer:
A)

Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Solve the following polynomial equation x^5-13x^3+36x=0
One of the following is the correct answer. Which one?
A) x=0, x=+or-2, x=+or-3
B) x=1, x=2, x=18
C) x=+or-2, x+or-3
D) x=+or-3, x=+or-5
E) x=0, x=+or-3, x=+or-12
Factor out GCF, x to get: 
               x[x2(x2 - 9) - 4(x2 - 9)] = 0 
                         x(x2 - 4)(x2 - 9) = 0
          x(x - 2)(x + 2)(x - 3)(x + 3) = 0 
                                                <===== CHOICE A)

Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

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:
, ,
which is the shorthand way of saying
x = 0, x = -2, x = 2, x = -3, x = 3

Answer: Choice A

--------------------------------------------------------

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.

RELATED QUESTIONS

Solve for x the inequality equation (x+9)/(x+1) -2 > 0 The correct answer is one of... (answered by greenestamps,math_tutor2020)
Solve the following equation.... (answered by ankor@dixie-net.com)
which of the following is not a solution to the equation: F(x) = x^5 + 5x^3 - 36x? a) (answered by MathLover1)
When asked to factor completely a certain polynomial, four students gave the following... (answered by Alan3354,College Student)
Which is the polynomial with integer coefficients that has degree 3 with zeros 0, -2, and (answered by mccravyedwin)
Pls help! If you can answer either one of them, I am so appreciated. 1.(x-2)(x+3)^2 /... (answered by Fombitz)
Pls help! If you can answer either one of them, I am so appreciated. 1.(x-2)(x+3)^2 /... (answered by Fombitz)
Solve the equation, x^2+2x=-17 One of the following answers is correct. Which one is... (answered by ikleyn,Alan3354)
Pls help! If you can answer either one of them, I am so appreciated. 1.(x-2)(x+3)^2 /... (answered by Fombitz)