• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

[Urgent] I have a polynomial, and I need to find a and b but don't know how! (1 Viewer)

GrizzlyPwnage

New Member
Joined
Aug 27, 2014
Messages
10
Gender
Male
HSC
2015
This is the question and I dont know how to do it, any feedback would be appreciated.

When the polynomial 2x3+ax2+bx+1 is divided by x-1 the remainder is 4; when divided by x-2, the remainder is 21. Find the values of a and b.


[Edit]
I also have a second question which is similar, but I am still having trouble with.

Given that (x+1) and (x-2) are two factors of x4-3x3+ax2+bx+6, find a and b and the two other factors.
 
Last edited:

seventhroot

gg no re
Joined
Aug 3, 2014
Messages
2,803
Gender
Male
HSC
2013
Re: [UrgentI have a polynomial, and I need to find a and b but don't know how!

sub in x = 1 and get an equation in a and b and equate that to 4

sub in x = 2 and get another equation in a and b and equate that to 21

solve equations simultaneously
 

GrizzlyPwnage

New Member
Joined
Aug 27, 2014
Messages
10
Gender
Male
HSC
2015
Re: [UrgentI have a polynomial, and I need to find a and b but don't know how!

Thankyou so much, I have my prelim exam on Tuesday and this really helps.
 

GrizzlyPwnage

New Member
Joined
Aug 27, 2014
Messages
10
Gender
Male
HSC
2015
Re: [UrgentI have a polynomial, and I need to find a and b but don't know how!

When you say "equate that to 4" what do you mean by that?
 

seventhroot

gg no re
Joined
Aug 3, 2014
Messages
2,803
Gender
Male
HSC
2013
Re: [UrgentI have a polynomial, and I need to find a and b but don't know how!

sub everything in and then = 4
 

GrizzlyPwnage

New Member
Joined
Aug 27, 2014
Messages
10
Gender
Male
HSC
2015
Re: [UrgentI have a polynomial, and I need to find a and b but don't know how!

Okay thankyou, your help is much appreciated.
 

seventhroot

gg no re
Joined
Aug 3, 2014
Messages
2,803
Gender
Male
HSC
2013
Re: [UrgentI have a polynomial, and I need to find a and b but don't know how!

nws :D
 

GrizzlyPwnage

New Member
Joined
Aug 27, 2014
Messages
10
Gender
Male
HSC
2015
If anyone would be able to help me with my second question It would be appreciated, thankyou.
 

seventhroot

gg no re
Joined
Aug 3, 2014
Messages
2,803
Gender
Male
HSC
2013
If anyone would be able to help me with my second question It would be appreciated, thankyou.
it's the same thing as the first one and then divide use polynomial division to find the other factors

IMO; you really need to work on your problem solving skills. you need to be able to read the question -> abstract it -> solve the problem
 

GrizzlyPwnage

New Member
Joined
Aug 27, 2014
Messages
10
Gender
Male
HSC
2015
Back again today, Im having trouble with my polynomials in general so sorry for all the questions.

The polynomial f(x)= ax2+bx+c has zeros 4 and 5, and f(-1) = 60. Evaluate a,b and c.
 

seventhroot

gg no re
Joined
Aug 3, 2014
Messages
2,803
Gender
Male
HSC
2013
what have you attempted thus far? there is no point me typing all the answers and you doing nothing
 

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
Back again today, Im having trouble with my polynomials in general so sorry for all the questions.

The polynomial f(x)= ax2+bx+c has zeros 4 and 5, and f(-1) = 60. Evaluate a,b and c.
This is only a quadratic. Fact it has zeros 4 and 5 means:

f(x) = a(x-4)(x-5)

f(-1) = a(-1-4)(-1-5) = 60 ==> 30a = 60 ==> a = 2

So f(x) = 2(x-4)(x-5) = 2x^2 - 18x + 40

So: a=2, b=-18 and c=40
 

seventhroot

gg no re
Joined
Aug 3, 2014
Messages
2,803
Gender
Male
HSC
2013
This is only a quadratic. Fact it has zeros 4 and 5 means:

f(x) = a(x-4)(x-5)

f(-1) = a(-1-4)(-1-5) = 60 ==> 30a = 60 ==> a = 2

So f(x) = 2(x-4)(x-5) = 2x^2 - 18x + 40

So: a=2, b=-18 and c=40
this is a nice solution. Personally I didn't see that and would have set up a Matrix
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top