Finding constants of a quartic equation

  • Thread starter chwala
  • Start date
  • #1
chwala
Gold Member
748
59

Homework Statement:

The polynomial ##4x^4+Ax^2+11x+b## where ##a## and ##b## are constants, is denoted by ##p(x)##. It is given that ##p(x)## is divisible by ##x^2-x+2##. Find the values of ##a## and ##b##

Relevant Equations:

factor theorem
##4x^4+Ax^2+11x+b##≡##(x^2-x+2)(ax^2+mx+d)##
≡##ax^4+(m-a)x^3+(d-m+2a)x^2+(2m-d)x+2d##

##a=4##
##m-a=0, m=4##
##d-m+2a=A##
##2m-d=11##
##2d=b##

##A=1##, ## B=-6 ## is that correct?
 

Answers and Replies

  • #2
ehild
Homework Helper
15,521
1,898
Homework Statement:: The polynomial ##4x^4+Ax^2+11x+b## where ##a## and ##b## are constants, is denoted by ##p(x)##. It is given that ##p(x)## is divisible by ##x^2-x+2##. Find the values of ##a## and ##b##
Relevant Equations:: factor theorem

##4x^4+Ax^2+11x+b##≡##(x^2-x+2)(ax^2+mx+d)##
≡##ax^4+(m-a)x^3+(d-m+2a)x^2+(2m-d)x+2d##

##a=4##
##m-a=0, m=4##
##d-m+2a=A##
##2m-d=11##
##2d=b##

##A=1##, ## B=-6 ## is that correct?
Yes.
 
  • #3
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,317
1,007
@chwala :
Try to be consistent with labeling of parameters.
In the given quartic polynomial,you use ##A## and ##b##. The question then asks for values of ##a## and ##b##. Then in the unknown quadratic factor, you use ##a## which is all together different from previous ##a## or ##A##, although, by inspection you can determine that this ##a=4## .

In the answer, you have ##A## and ##B##.
 
  • #4
chwala
Gold Member
748
59
That was a pretty easy question.....i guess i was just over thinking on it.... :biggrin: :cool:
 
  • #5
chwala
Gold Member
748
59
@chwala :
Try to be consistent with labeling of parameters.
In the given quartic polynomial,you use ##A## and ##b##. The question then asks for values of ##a## and ##b##. Then in the unknown quadratic factor, you use ##a## which is all together different from previous ##a## or ##A##, although, by inspection you can determine that this ##a=4## .

In the answer, you have ##A## and ##B##.
true, i had seen that but i just ignored ........noted
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,317
1,007
(Using uppercase ##A## and ##B## for the problem), I suppose you could say, if ##x^2-x+2## is a factor of ##4x^4+Ax^2-11x+B##, then by inspection, the other factor must be of the form ##4x^2+mx+\frac B 2 ##.

Multiplying the quadratics gives:
##\left(x^2-x+2\right) \left(4x^2+mx+\frac B 2 \right) = 4x^4+(m-4)x^3+\left(8-m+\frac B 2 \right)x^2 +\left(2m-\frac B 2\right)x +B##.

Match coefficients.

etc.
 
Last edited:
  • Like
Likes chwala
  • #7
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,259
5,836
Homework Statement:: The polynomial ##4x^4+Ax^2+11x+b## where ##a## and ##b## are constants, is denoted by ##p(x)##. It is given that ##p(x)## is divisible by ##x^2-x+2##. Find the values of ##a## and ##b##
Relevant Equations:: factor theorem

##4x^4+Ax^2+11x+b##≡##(x^2-x+2)(ax^2+mx+d)##
≡##ax^4+(m-a)x^3+(d-m+2a)x^2+(2m-d)x+2d##

##a=4##
##m-a=0, m=4##
##d-m+2a=A##
##2m-d=11##
##2d=b##

##A=1##, ## B=-6 ## is that correct?
Not sure it's any easier, but another method is to use ##x^2=x-2## to substitute for ##x^2##, first in the ##x^4## term, then in the resulting ##x^2## term.
The resulting linear expression must equal zero.
 
  • Like
Likes SammyS and etotheipi
  • #8
chwala
Gold Member
748
59
Not sure it's any easier, but another method is to use ##x^2=x-2## to substitute for ##x^2##, first in the ##x^4## term, then in the resulting ##x^2## term.
The resulting linear expression must equal zero.
i do not get this approach ##x^2=x-2## how? from where...
 
  • #9
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,259
5,836
i do not get this approach ##x^2=x-2## how? from where...
If p(x) is divisible by ##x^2−x+2## then every solution of ##x^2−x+2=0## makes p(x) zero.
 
  • Like
Likes chwala and etotheipi
  • #10
chwala
Gold Member
748
59
aaaaaah....seen that..thanks haruspex
 

Related Threads on Finding constants of a quartic equation

Replies
15
Views
322
Replies
5
Views
668
Replies
11
Views
1K
Replies
5
Views
4K
Replies
18
Views
3K
  • Last Post
Replies
1
Views
3K
  • Last Post
3
Replies
62
Views
5K
  • Last Post
Replies
12
Views
664
  • Last Post
Replies
2
Views
959
  • Last Post
Replies
1
Views
1K
Top