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

Could someone please explain the above again?

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

We have been told that :

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

Or \(x^{4}-1 = x^{4}+x^3c+x^2d+ax^3+acx^2+adx+bx^2+bcx+bd \to \\
\\
x^{4}-1 = x^{4}+x^3(a+c)+x^2(b+d+ac)+x(ad+bc)+bd\)

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.

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 ?

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

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

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.

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!

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

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