remainder theorem problem: p(x) = x^6 + ax^4 + bx^2 + c; a,b,c integers; p(7) = 0, p(1+sqrt(4))=0, find p(2)

Your example => a, b
My problem=> a,b,c
But there are only 2 equations.
 
topsquark said:
If a,b, and c are integers, you are supposed to know that if [imath]g+\sqrt{h}[/imath] is a solution, then so is [imath]g-\sqrt{h}[/imath] . You do have three equations.

-Dan
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I think this concept is used only when g+root h
g is rational number and h is irrational number

why p(7) = p(5+2) can we get p(5-2)
or p(7) = p(4+3) can we get p(4-3)
 
juandiaz said:
I think this concept is used only when g+root h
g is rational number and h is irrational number

why p(7) = p(5+2) can we get p(5-2)
or p(7) = p(4+3) can we get p(4-3)
Click to expand...
Why do you want to query what you've been told and come up with some outlandish extrapolation?
You've been given the three factors of the polynomial so you can construct three equations to solve for a, b & c and thence evaluate p(2).
Please go ahead and do so.
 
I’ve just known. This problem i wrote was incorrect.
I will re-write again.
 
juandiaz said:
I think this concept is used only when g+root h
g is rational number and h is irrational number

why p(7) = p(5+2) can we get p(5-2)
or p(7) = p(4+3) can we get p(4-3)
Click to expand...
Yes, I should have been more specific here.

Specifically, it holds when h is not a perfect square.

But the point still stands: you have three equations. Can you finish?

-Dan
 
I agree with your answer to the corrected problem, though you failed to explain some steps.

I think what you are saying is that because the coefficients are integers, [imath]1+\sqrt{3}[/imath] being a root implies that its surd conjugate, [imath]1-\sqrt{3}[/imath], is also a root; and because the function is even, their negatives are also roots. Given that, your work is correct.

I took a slightly different path, treating p(x) as q(x^2), so that the square of each root pf p is a root of q; but otherwise, my work is similar.

Once the problem was copied correctly, this wasn't so hard.
 
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stapel said:
Where are you getting that [imath]p(-7)[/imath] is also equal to zero? Where are you getting that [imath]p\left(-1+\sqrt{3\;}\right)[/imath] is also equal to zero?

Where are you getting that [imath]-1 \pm \sqrt{3\;}[/imath] are factors?

Are you familiar at all with the Quadratic Formula?
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Because P(x) has only even degree
x^6, x^4, x^2 so we can get p(-x) = p(x).
So p(-7)=p(7)
And from p(1+root3)=0 we get p(-1-root3)

And another concept : an answer (sum of rational and irrational) has conjugate answer
to produce coefficient that are rational.
So from p(1+root3)=0 we get p(1-root3)=0
and from p(-1-root3)=0 we get p(-1+root3)=0.
 
juandiaz said:
Because P(x) has only even degree
x^6, x^4, x^2 so we can get p(-x) = p(x).
Click to expand...
Because P(x) has only even degrees
x^6, x^4, x^2 AND c=cx^0, so we get p(x)=p(-x).

You were lucky that all the degrees were even (even the constant had an even power of x).
What if you had, P(x) = x^5 + x^3 + x^1 + 6?
Would you say because P(x) has only odd degrees that p(x) is an odd function so p(-x) = -p(x)? This would NOT be true. In this example, P(x) has both odd and even degrees! 6=6x^0 and 0 is NOT odd (it is even!)

All of your work was good but not saying that the constant c is even was a big mistake. I would have graded your work harshly because of that. You need to know that a constant is considered even with respect to the power of x!!!
 
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