Math Is Fun Forum

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Algebra Problems

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.

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?

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.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra Problems

It was all ironed out for you over in another thread.

---

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

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra Problems

It is not necessary. You have the correct answer. Until someone can come up with another polynomial that works too why bother?

---

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.

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.

O.K. It is g(x)/f(x) My comments were for f(x)/g(x). g(x) has to be divisible by f(x). All the zeroes of f(x) should be there in g(x) In addition a/b that you have taken is to be integer.That is my opinion.

---

{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

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.

1|	a	b	1-a-b	0	0	-1
		a	a+b	1	1	1
	………………	………………	………………	………………	………………	………………
	a	a+b	1	1	1	0

1|	a	`-3-a	1	1	1
		a	-3	-2	-1
	………………	………………	………………	………………	………………
	a	-3	-2	-1	0 

---

{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

Thank you! I did it the same way!

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Algebra Problems

Hi dazzle1230,
According to your earlier comment, you substituted x=y+1 ,expanded and got 3 simultaneous equations in a,b,c. It is o.k. when you are not writing exam or if you are allowed to use programmable calculators but it is awfully simple if you can solve them one by one.

---

{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.}