a, b, c, and d are constants such that: a, b, and c are all

Question

a, b, c, and d are constants such that: a, b, and c are all smaller than d; a, c, and d are all larger than b; and x^4 – 1 = (x^2 + ax + b)(x^2 + cx + d) for all numbers x. What is the sum of a and c?

A. –4
B. –1
C. 0
D. 1
E. 4

A. –4
B. –1
C. 0
D. 1
E. 4

GyanOne · Accepted Answer

x4 – 1 = (x2 + ax + b)(x2 + cx + d)
=> x^4 - 1 = x^4 +(c+a)x^3+ (b+d+ac)x^2+ (ad+bc)x + bd

Since this is an identity that holds for all x, we can equate the coefficients
=> a+c=0 (equating the coefficients of x^3)

Choice C

kashishh · Answer

Dear Gyanone,
cud you pls elaborate on how these coefficients are equated?
what is the underlying concept?

audiogal101 · Answer

Could someone please explain the above again?

papua · Answer

GyanOne if a+c=0, then a=-c and we receive that : x^4 - 1=x^4+(b+d-c^2)x^2+(-cd+bc)x+bd. How can we say that this equation is correct ? Thanks for further explanation

mau5 · Answer

We have been told that :

x^{4}-1 = (x^2 + ax + b)(x^2 + cx + d)  for ALL values of x.

Or  x^{4}-1 = x^{4}+x^3c+x^2d+ax^3+acx^2+adx+bx^2+bcx+bd 	o

x^{4}-1 = x^{4}+x^3(a+c)+x^2(b+d+ac)+x(ad+bc)+bd

or  x^3(a+c)+x^2(b+d+ac)+x(ad+bc)+bd+1 = 0  for all values of x.

This means that the above equation equals zero for x=1, x=0, x=-5 or any other value of x. However, this is a polynomial of 3rd degree, that is, it can have at most 3 roots. Thus, as for any value of x, the above equation is equal to zero, hence, all the co-efficients of the equation must be individually equal to zero.Thus, a+c = 0

The question could have asked what is the value of bd also.

C.

papua · Answer

mau5 thanks for the explanation.

you mention that "The question could have asked what is the value of bd also. "

How we guess what is the value of bd ?

mau5 · Answer

Because all the co-efficients are zero, we have bd+1=0, bd=-1.

papua · Answer

Dear mau5 for bd to equal -1, b should be -1/d, or vice versa.

Also if bd=-1 and a=0;c=0, x^3(a+c)+x^2(b+d+ac)+x(ad+bc)+bd+1=0 => x^2((-d^2-1)/d)=0

and this is only true if x=0 and not for all numbers x.

Can you please explain further ?

Thanks in advance

mau5 · Answer

There is no reason to tread down this path, as all the co-efficients are already zero. Also, you have for no reason assumed that a=c=0.

All we know is :

a+c = 0,

b+d+ac = 0,

ad+bc = 0,

and bd=-1.

iliavko · Answer

Hi everyone,

In these "valid for all X" cases, do we really have to unfold the entire equation? Like I had no trouble to determine bd=-1 just assume X=0

But then, is there a faster way to determine A and C ? Deadmaus's way seems too long and full of traps, like with the time pressure, fatugue etc, it begs for a miskate. Or I am wrong about it?

Is it possible to use the bd=-1 information to determine A and C? There is DS quesiton, equating-coefficients-in-a-quadratic-equation-103919.html basically by assuming x=0 and then substituting and then assuming x=1 you simplify your life a lot, I know the questions are slightly different, but is it possible to apply the same principles here?

Also, I didn't see that the "a, b, and c are all smaller than d; a, c, and d are all larger than b" parte of the question stem being used to answer the question. we get b < a,c < d, I am I right?
Can we use this info plus the fact that bd=-1 to directly answer the question without the hairy unfolding and factoring of x^4 – 1 = (x^2 + ax + b)(x^2 + cx + d) ?

Thank you a lot for the time!

mdacosta · Answer

I'm still not following why every coefficient must equal zero (or how you got X=0/X=-5), sorry. I understand for x^4 - 1 = equation that if x=1 then the RHS must = 0 as well, but with the +1 at the very end it seems to be the equation if x=0 would go 0 + 0 + 0 + 0 + 1 = 0 which doesn't work.

Brunel or Mau5 could you kindly elaborate for us slower folks? Thanks!

Michael

bumpbot · Answer

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a, b, c, and d are constants such that: a, b, and c are all

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