# Finding constants of a quartic equation

Gold Member

## 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?

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ehild
Homework Helper
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.

SammyS
Staff Emeritus
Homework Helper
Gold Member
@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##.

Gold Member
That was a pretty easy question.....i guess i was just over thinking on it....  Gold Member
@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

SammyS
Staff Emeritus
Homework Helper
Gold Member
(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 ##.

##\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:
• chwala
haruspex
Homework Helper
Gold Member
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.

• SammyS and etotheipi
Gold Member
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...

haruspex
Homework Helper
Gold Member
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.

• chwala and etotheipi
Gold Member
aaaaaah....seen that..thanks haruspex