If all roots of the polynomial x^4 - 2x^3 - 9x^2 + 2x + 8

First of all ccax is a math Guru and his Qs are definitely harder than GMAT Qs . Moreover, I don't think we need to know 4th degree polynomail solutions for GMAT..

It is easy to conclude that x = 1 and x = -1 satisfy the equation. Put x=1/-1, and the expression becomes 0.

Now the hard part is factorizing the polynomail. You need to know a technique called synthetic division. I somehow remember it from my high school days.. You can check out some websites like mathworld or purplemath... and learn it, but I think it is really not necessary for GMAT. PVUE will give you such Qs only when you are scoring above 60 in Q.

ccax is there an easier way to solve this Q?

First of all giddi77, thanks for the flowers
I wish I could exchange the compliment into a decent GMAT score...

I don't think my questions are always harder than those of GMAT,
but I tried a few times to create a question that required to think
a bit different.

It's certainly true that you won't NEED 4th degree polynomials.
Because of that, I chose integers for the roots, so that the question
won't be too hard.

If you consider that a 4th degree polynomial can be represented as
(x-a)(x-b)(x-c)(x-d), where a, b, c and d are the roots, and you
expand this, then the only number independent from x is a*b*c*d,
which in our case was +8, representing the product of all roots of
the polynomial.

There aren't that many possibilites if all roots have to be integers,
so just try a few values. You won't need polynomial division in the
GMAT.

In calculating the roots of a polynomial, there are general formulas
for a polynomial up to 4th degree (that are exponentially growing
more complicate for 3rd and especially 4th degree). For polynomials
of a degree bigger than 4, there are only formulas for special cases.
As for the GMAT, you only need to bear in mind the formula for
x^2 + b x + c = 0.

the way to solve this would be to compare the coefficients of the same exponent of x


so since it is a 4th root, there should be four roots, assume the roots to be a, b, c, d

Thus, (x-a)(x-b)(x-c)(x-d) = the given equation.

Expand the LHS... you should get an expression with coefficients of (x^4), (x^3), (x^2), (x^1) and (x^0, ie, no x). Compare the LHS and RHS for the same powers of X.

You should get 4 equations!
Solve them to get the values of a,b,c,d!

They will the four roots of the given equations.

Don't worry buddy! You will soon have one that you might not want to exchange!

Thanks for explantion Sir! I agree that sometimes we need to think too.. Just applying formulas and getting some results is not solving problems!

Once you find that x = +/-1 work, the best way is to use:
(x-1)(x+1)(ax^2+bx+c) ... expand and compare the co-efficients, to get a,b,c... and c = -8 can be directly concluded.

You're right with your suggestion. But just to give you an impression
of the complexity of the formula for the first of the four solutions of
a 4th degree polynomial, hava a look at the image. You can be
sure that you won't encounter this on GMAT.

x^4 + a x^3 + b x^2 + c x + d = 0

One of the roots looks like (the others are similar):