Please help me with this problem.

Prove that if p(x) is polynomial with integer coefficients and p(√2)=0 then p( -√2 )=0.

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

david mcmillan see this

Same thing Sandeep Bhardwaj said

U Z - 6 years, 4 months ago

But $p(2) = 0$ doesn't imply that $p(-2) = 0$ . How else should I be reading " $p(2)$ then $( -2 ) = 0$ " ?

Brian Charlesworth - 6 years, 4 months ago

Sir he meant - $p(\sqrt{2})$

U Z - 6 years, 4 months ago

Sorry for the typo. I meant $p(-\sqrt{2})$

david mcmillan - 6 years, 4 months ago

@David Mcmillan – O.k., thanks for the update. However, the fact that $p(2) = 0$ won't imply that $p(-\sqrt{2}) = 0$ either. On the other hand, if $p(x)$ has integer coefficients then $p(\sqrt{2}) = 0 \Leftrightarrow p(-\sqrt{2}) = 0$ . This comes as a result of the irrational conjugate roots theorem, which states that for any polynomial $p(x)$ with rational coefficients, if $a + b\sqrt{c}$ is a root of $p(x)$ , where $a,b$ are rational and $\sqrt{c}$ is irrational, then $a - b\sqrt{c}$ must also be a root of $p(x)$ .

To save some time, I'm just going to give you a link to a proof of the general theorem, (which was surprisingly hard to find). Scroll down to theorem 16 on page 14 for the desired proof. You can adapt this proof for the special case $a = 0, b = 1$ to simplify it a bit. Good question; I don't think I've ever come across a proof of this theorem before. :)

Brian Charlesworth - 6 years, 4 months ago

@Calvin Lin @brian charlesworth @Sreejato Bhattacharya Please help.

david mcmillan - 6 years, 4 months ago

@Sudeep Salgia @Sandeep Bhardwaj Also try and help

david mcmillan - 6 years, 4 months ago

This is the basic property of a polynomial equation. In fact, this is true for all rational coefficients. You try to prove it by taking a quadratic equation and then try to generalize that.

Sandeep Bhardwaj - 6 years, 4 months ago

Do you mean that if $p(2) = 0$ then $p(-2) = 0$ ? If so, then this is not the case. If $p(x) = x^{2} - x - 2$ then $p(2) = 0$ but $p(-2) = 4$ . Perhaps you meant something else with the terminology "( -2 ) = 0"?

What can be said is that if a complex number $z$ is a root of a polynomial then so is its conjugate $\bar {z}$ , i.e., $p(z) = 0 \Leftrightarrow p(\bar {z}) = 0$ .

Brian Charlesworth - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$