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How would you find all the solutions of x^5 - 3x^4 - 2x^3 - 4x^2 - 24x + 32

Guest Mar 26, 2017
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 #1
avatar+17601 
+1

To find rational answers, factor the constant term (32) and divide each of these numbers by the factors of the coefficient of the iinitial term (1).

The factors of 32 are +1, -1, +2, -2, +4, -4, +8, -8. +16. -16. +32, and -32.

Try these, one at a time, until you find one that makes the expresssion zero.

If you try +1, the expression becomes zero; this tells you that one of the answers is +1, and one of the factors of the expression is (x - 1).

Now, there are a couple of things that you can do.

You can divide the original expression by (x - 1) to give you a smaller expression:

         x4 - 2x3 -4x2 - 8x - 32

Again, factor the constant term (32) and divide each of these numbers by the facotrs of the coefficient of the initial term (1).

And again, these factor are +1, -1, +2, -2, +4, -4, +8, -8. +16. -16. +32, and -32.

Try these, one at a time, until you find one that makes the expression zero.

This will give you a second solution ... create the factor .... divide, and continue this process until you can no longer get another solution.

geno3141  Mar 26, 2017
 #2
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+1

Solve for x:
x^5 - 3 x^4 - 2 x^3 - 4 x^2 - 24 x + 32 = 0

The left hand side factors into a product with four terms:
(x - 4) (x - 1) (x + 2) (x^2 + 4) = 0

Split into four equations:
x - 4 = 0 or x - 1 = 0 or x + 2 = 0 or x^2 + 4 = 0

Add 4 to both sides:
x = 4 or x - 1 = 0 or x + 2 = 0 or x^2 + 4 = 0

Add 1 to both sides:
x = 4 or x = 1 or x + 2 = 0 or x^2 + 4 = 0

Subtract 2 from both sides:
x = 4 or x = 1 or x = -2 or x^2 + 4 = 0

Subtract 4 from both sides:
x = 4 or x = 1 or x = -2 or x^2 = -4

Take the square root of both sides:
Answer: |x = 4   or x = 1   or x = -2   or x = 2 i   or x = -2 i

Guest Mar 26, 2017
 #3
avatar+7055 
+1

How would you find all the solutions of x^5 - 3x^4 - 2x^3 - 4x^2 - 24x + 32

 

\( x^5 - 3x^4 - 2x^3 - 4x^2 - 24x + 32 \)     is a term. There is no solution.

 

I guess you mean a function and you're looking for the zeros.

 

\(f(x)= x^5 - 3x^4 - 2x^3 - 4x^2 - 24x + 32 =0\)

 

Solution guess and simplify term several times by polynomial division.

 

\(x_1=-2\)

 

\(x_2=1\)

 

\(x_3=4\)

 

laugh  !

asinus  Mar 26, 2017

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