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## #1 Help Me ! » Algebra Problems » 2016-12-08 13:07:01

dazzle1230
Replies: 4

1)If f is a polynomial of degree 4 such that f(0) = f(1) = f(2) = f(3) = 1 and f(4)=0, then determine f(5).
2)Let f(x) be a quartic polynomial with integer coefficients and four integer roots. Suppose the constant term of f(x) is 6.  Is it possible for x=3 to be a root of f(x)?  Is it possible for x=3 to be a double root of f(x)?

## #2 Re: Help Me ! » Algebra Problems » 2016-12-05 11:34:20

Thank you! I did it the same way!

## #3 Re: Help Me ! » Algebra Problems » 2016-12-04 10:25:23

I see the thread...but do we take in account the multiplicity of roots?

## #4 Re: Help Me ! » Algebra Problems » 2016-12-04 08:10:37

dazzle1230 wrote:

and for part one of the second problem, I said that g(x) has to be divisible by f(x):
g(x)=a(x+5)(x+3)(x-2)(x-4)(x-8)
f(x)=b(x+3)(x-4)(x-8)
Once we divide f(x) from g(x) and simplify the quadratic, we get that (a/b)x^2+3x10...is that correct?
If so, how do we proceed part 2 of the problem?

I do not know the technicalities but is it not necessary that a has to be divisible by b also?
In my opinion the coefficient of highest degree in f(x) must be divisible by the coefficient of highest degree in g(x) and f(x) should contain all the zeroes of g(x) ,even the repeated ones.

So are you saying that g(x) doesnt have to be divisible by f(x)?  Should we take multiplicity of roots into account?  Also, for part 2, I'm thinking that the degree of f(x) has to be 3 and the degree of g(x) has to be 5, using the fundamental theorem of algebra, so there won't be multiplicity of roots.

## #5 Re: Help Me ! » Algebra Problems » 2016-12-04 07:45:54

dazzle1230 wrote:

Yes, I got 6x^5-15x^4+10x^3
Thanks!

Can you illustrate the method in detail?

First I knew that f(x) had to be in form ax^5+bx^4+cx^3 in order to be divisible by x^3.  I let y=x-1, so x=y+1.  So then f(x)-1 is only divisible by (x-1)^3 if f(y+1)-1 is divisible by (x-1)^3.  After expanding, I solved a system of equations to find a, b, and c.

## #6 Re: Help Me ! » Algebra Problems » 2016-12-03 10:12:46

I'm not sure how to do part 2 though. Do we need the fundamental theorem of algebra?

## #7 Re: Help Me ! » Algebra Problems » 2016-12-03 06:07:26

Is my reasoning for the second problem correct?

## #8 Re: Help Me ! » Algebra Problems » 2016-12-03 05:32:13

The first problem or the second one?

## #9 Re: Help Me ! » Algebra Problems » 2016-12-03 05:29:11

and for part one of the second problem, I said that g(x) has to be divisible by f(x):
g(x)=a(x+5)(x+3)(x-2)(x-4)(x-8)
f(x)=b(x+3)(x-4)(x-8)
Once we divide f(x) from g(x) and simplify the quadratic, we get that (a/b)x^2+3x10...is that correct?
If so, how do we proceed part 2 of the problem?

## #10 Re: Help Me ! » Algebra Problems » 2016-12-03 05:26:15

Yes, I got 6x^5-15x^4+10x^3
Thanks!

## #11 Re: Help Me ! » Algebra Problems » 2016-12-02 14:29:59

How would we use the fundamental theory of algebra to prove problem 2?

## #12 Re: Help Me ! » Algebra Problems » 2016-11-29 11:40:02

Thank you!  I also have two other problems I need help on:
1) Find a polynomial f(x) of degree 5 such that both of these properties hold: f(x)-1 is divisible by (x-1)^3 and f(x) is divisible by x^3
2)
Part 1:
Let f(x) and g(x) be polynomials.
Suppose f(x)=0 for exactly three values of x: namely, x=-3,4, and 8.
Suppose g(x)=0 for exactly five values of x: namely, x=-5,-3,2,4, and 8.
Is it necessarily true that g(x) is divisible by f(x)? If so, carefully explain why. If not, give an example where g(x) is not divisible by f(x).

Part 2:
Generalize: for arbitrary polynomials f(x) and g(x), what do we need to know about the zeroes (including complex zeroes) of f(x) and g(x) to infer that g(x) is divisible by f(x)?

## #13 Help Me ! » Algebra Problems » 2016-11-26 11:49:26

dazzle1230
Replies: 33

1) Find all integers

for which
is an integer.
2) Find the range of the function
Enter your answer in interval notation. (I got [3,5] for this problem, and I also graphed it, but the answer isn't right)
3) Suppose the polynomial f(x) is of degree 3 and satisfies f(3)=2, f(4)=4, f(5)=-3, and f(6)=8.  Determine the value of f(0).

yup, I understand that!

Thank you so much!

dazzle1230
Replies: 10

Circle

intersects the hyperbola
at
, and two other points. What is the product of the y coordinates of the other two points?

## #17 Re: Help Me ! » Algebra Problems » 2016-11-12 05:56:24

It was actually a>0, sorry

Is that for proving a<-4?

## #18 Re: Help Me ! » Algebra Problems » 2016-11-11 14:46:15

I proved that a>0, but I'm having trouble proving that a<-4

## #19 Re: Help Me ! » Algebra Problems » 2016-11-10 12:20:57

I need help on 3:

I know that a>-1, the product of the roots is greater than 1, and the discriminant is greater than 0.  That's about it

## #20 Re: Help Me ! » Algebra Problems » 2016-11-10 06:46:20

Thank you! Its right now.

## #21 Re: Help Me ! » Algebra Problems » 2016-11-09 13:39:47

The sum would be 9.

9 isn't right though

I still don't know how to proceed the 3rd question though

## #22 Re: Help Me ! » Algebra Problems » 2016-11-09 10:06:49

I think the only roots are

,
,
, and

I still don't understand the 3rd question with Vieta's formula either

## #23 Re: Help Me ! » Algebra Problems » 2016-11-08 12:51:23

Thanks for the other hints!  But I still can't figure out 2.

\frac 1{x_1}+\frac 1{x_2}<1  is

## #24 Help Me ! » Algebra Problems » 2016-11-07 11:08:25

dazzle1230
Replies: 19

1) What is the smallest distance between the origin and a point on the graph of

2) Determine the sum of all real numbers x satisfying (x^2-4x+2)^{x^2-5x+2}=1.

3)Let a, b be real numbers, and let x_1, x_2 be the roots of the quadratic equation x^2+ax+b=0.  Prove that if x_1, x_2 are real and nonzero, \frac 1{x_1}+\frac 1{x_2}<1, and b>0, then |a+2|>2.

Only have a lead on #2:
I let y=x^2-4x+2 so that the equation would read y^{y-x}=1. Since y^0=1 and they have the same base, I knew that y=x. Substituting again, I would get that x^2-4x+2=x. Subtracting x from both sides, I then used the quadratic formula to find the roots of x. But once I added the roots together, the answer was wrong. Can anyone tell me what I'm doing wrong? (I got 5.)

## #25 Re: Help Me ! » Help! » 2016-11-03 10:18:53

I see now.  It's 1/4