Polinomial proof

[Guest]

Hi,
I need help with this task.
We've got polinomial W(x). Proof that if [tex]x- \sqrt(2) | W(x)[/tex] then [tex]x^2 -2 | W(x)[/tex].
Thanks for help,
John

[Mathmaven53]

Let W(x) be the polynomial. Assume it has integer coefficients.
Divide W(x) by x^2 - 2

W(x) = (x^2 - 2)Q(x) + a x + b
where Q(x) is the quotient polynomial and has integer coefficients and a and b are integers

since x - sqrt 2 divides W(x) then W(sqrt 2) = 0

So W(sqrt 2) = a sqrt 2 + b = 0

Now if a not equal to zero

sqrt 2 = -b/a
which would mean that sqrt 2 is a rational number.
This is false.

So the assumption that a not equal to zero leads to a contradiction

So a = 0

Then from a sqrt 2 + b = 0

b = 0

W(x) = (x^2 - 2) Q(x)

Thus W(x) divisible by x^2 - 2

[Guest]

You CAN'T prove it! It is NOT true!

For example, [tex]W(x)= x- \sqrt{2}[/tex] is clearly divisible by [tex]x- \sqrt{2}[/tex] but not by [tex]x^2- 2[/tex].

What is true is the other way around: If W(x) is divisible by [tex]x^2- 2[/tex] then it is divisible by [tex]x- \sqrt{2}[/tex] because [tex]x- \sqrt{2}[/tex] is a factor of [tex]x^2- 2[/tex].

What is also true, and perhaps what you intended, is "If W(x), a polynomial with integer coefficients is divisible by [tex]x- \sqrt{2}[/tex] then it is divisible by [tex]x^2- 2[/tex]".

[Angus53]

To solve this problem, we'll utilize the Factor Theorem, which states that if \( (x - c) \) is a factor of a polynomial \( W(x) \), then \( W(c) = 0 \).

Given that \( (x - 2) \) is a factor of \( W(x) \), we know that \( W(2) = 0 \).

Now, we want to prove that \( (x^2 - 2) \) is a factor of \( W(x) \). To do this, we'll show that \( W(\sqrt{2}) = 0 \) and \( W(-\sqrt{2}) = 0 \).

Since \( W(2) = 0 \), it implies that \( W(\sqrt{2}) \cdot W(-\sqrt{2}) = 0 \), according to the Conjugate Roots Theorem. Therefore, if \( W(\sqrt{2}) = 0 \), then \( W(-\sqrt{2}) = 0 \), and vice versa.

Thus, we've proved that \( (x^2 - 2) \) is a factor of \( W(x) \).