Factor Polynomial of Degree 5

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nasi112

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Solve the equation. Give reasons for each step.

[MATH]x^5 + 5x^3 - 8x - 40 = 0[/MATH]
I tried to use the rational root theorem to obtain the factors, but none of them worked.

According to Wolfram, the answer is

[MATH]x = 1.86426[/MATH]
I know that this is just an approximation.

Is there a way to find the exact solution with steps?
 
An approximation is the best that can be done given this quintic polynomial.

Double check the exponents ... if the equation were [MATH]x^5 + 5x^4 - 8x - 40 = 0[/MATH], it would be possible.
 
Are you sure that you copied the problem correctly.

[MATH]x^5 + 5x^3 - 8x^2 - 40 = 0 \implies x^3(x^2 + 5) - 8(x^2 + 5) = 0 \implies[/MATH]
[MATH](x^3 - 8)(x^2 + 5) = 0.[/MATH]
That equation has only one real solution, namely x = 2.

[MATH]2^5 + 5 * 2^3 - 8 * 2^2 - 40 = \cancel {32} + 5 * 8 - \cancel {8 * 4} - 40 = 40 - 40 = 0. \ \checkmark[/MATH]
 
Thanks guys for the attempt. I wish if there is a mistype, but that was the exact problem!
 
HallsofIvy said:
Then x=1.86426 is a very poor approximation to 2!
Click to expand...
You are correct. 1.86426 is a very poor approximation of 2. Of course, 2 is not the correct answer. ?

[MATH]2^5 + 5 * 2^3 - 8 * 2 - 40 = 32 + 5 * 8 - 16 - 40 = 40 - 40 + 32 - 16 = 16 \ne 0.[/MATH]
 
nasi112 said:
Solve the equation. Give reasons for each step.

[MATH]x^5 + 5x^3 - 8x - 40 = 0[/MATH]
I tried to use the rational root theorem to obtain the factors, but none of them worked.

According to Wolfram, the answer is

[MATH]x = 1.86426[/MATH]
I know that this is just an approximation.

Is there a way to find the exact solution with steps?
Click to expand...
Roots of a quintic polynomial cannot be solved "exactly" using elementary algebra.
 
You should at least try the rational root theorem to check for rational roots.
 
It occurs to me that we have not answered the question posed for the actual equation given.

One way to find an answer is to note that f(0) is - 40 and f(2) = 16. Therefore, there must a root between 0 and 2. Try halfway.

f(1) = - 42. Therefore there must be a root between 1 and 2. Try halfway.

f(1.5) = - 27.53125. Therefore, there must be a root between 1.5 and 2. Try halfway.

f(1.75) is approximately - 11. Therefore there must a root between 1.75 and 2. Try halfway.

f(1.875) is approximately plus 1. Therefore there must be a root between 1.75 and 1.875.

Welcome to the world of numerical methods, which is a fancy word for systematic trial and error that computers are great at.

What I showed is the method of bisection (or binary search). My recollection is that there are far more efficient numerical methods for finding the roots of differentiable functions (for example, the Newton-Raphson method). But even though the problem is not solvable by purely algebraic methods, it is not actually a hard problem to solve to any degree of precision required.
 
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