Math Is Fun Forum

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Algebra Problems

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).

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Algebra Problems

For (3), no immediate ideas come to mind, except for plug in each point separately into

and solve the system.

EDIT: Doing so, we get the system as

Which has solutions

. Thus, the cubic is

and

.

I probably got this wrong though. Mind to check over?

Last edited by evene (2016-11-26 13:37:50)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra Problems

Hi;

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,624

Re: Algebra Problems

hi dazzle1230

Q2.  (-2,3) is certainly the lowest point.  Maybe it never reaches 5; just tends to it.  Try an open interval at the right hand end.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Algebra Problems

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)?

Thank you in advance!!!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra Problems

Hi;

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,624

Re: Algebra Problems

hi dazzle1230

Q2. Parts 1 and 2

Initially, I'm assuming real values only

Here's a possible example to consider:

If values can be complex then the (x² +2) and (x^4+7) terms would not exist since these would also give rise to zeros, and we're told the only zeros are as given.

But, for example, f could have a factor (x+3)^2 and still satisfy the requirements.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Algebra Problems

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Algebra Problems

hi dazzle1230

But, for example, f could have a factor (x+3)^2 and still satisfy the requirements.

Bob

Then f will have 4 factors.(-3 twice)

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

Bob

Administrator

Registered: 2010-06-20

Posts: 10,624

Re: Algebra Problems

Suppose f(x)=0 for exactly three values of x: namely, x=-3,4, and 8.

Number of factors and number of zeros are not the same thing.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra Problems

Hi;

I am forced to agree with bobbym on this one. His motto, "that even a blind squirrel finds an acorn once in a while," is right on this one. He has the polynomial answer, I saw it when I peeked at his work while he was eating. Is that unethical? Certainly not...the ole fellow would just say RIPOSTP.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Algebra Problems

Suppose f(x)=0 for exactly three values of x: namely, x=-3,4, and 8.

Number of factors and number of zeros are not the same thing.

Bob

I think they are 2 zeroes superposed one over the other.

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Algebra Problems

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

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Algebra Problems

Hi dazzle1230,
How did you solve prob (1) of  #5?

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Algebra Problems

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

Last edited by dazzle1230 (2016-12-03 06:29:31)

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Algebra Problems

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?

Last edited by dazzle1230 (2016-12-03 05:31:22)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra Problems

Hi;

That is correct!

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Algebra Problems

The first problem or the second one?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra Problems

This one.

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

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Algebra Problems

Is my reasoning for the second problem correct?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra Problems

The fact that Remainder equals zero means that f(x) | g(x).

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Algebra Problems

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

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra Problems

Part 2... I do not know how.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Algebra Problems

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.

Last edited by thickhead (2016-12-03 18:37:42)

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Algebra Problems

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

Can you illustrate the method in detail?

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}