what are these roots?

Question

q(x)= 2x^3+4x^2+7x+5
and another question: a=x b=-1 what polynomial is (x-1)^6 equal to?

Answer 1

There is only one zero (or root) in this equation.

q(x) = 2x^3 + 4x^2 + 7x + 5
= (x + 1) ( 2x^2 + 2x + 5)

Thus: The zero or x-intercept is at -1.

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What polynomial is (x-1)^6 equal to?
(x - 1)^6 = (x - 1) (x - 1) (x - 1) (x - 1) (x - 1) (x - 1)

= x^6 - 6x^5 + 15x^4 - 20x^3 +15x^2 - 6x +1

I don't understand what you mean by a = x, b = -1?
(Thank you dnnyo - I think I get it now)

(a+b)^6 =
a^6 + 6a^5b + 15a^4 b^2 + 20a^3 b^3 +15a^2 b^4 + 6a b^5 + b^6

Hopefully, this helps.
Good luck in your studies.

Answer 2

With cubics (at least the ones they give you at school), the trick is to spot one solution and then work out the others (if any). It's usually best to try 1, -1, 2, -2 etc. first - and if the remainders are getting way off, use the ones you've calculated already to help you guess where the solution is.

In this case, if you're alert you'll notice that sums of alternate coefficients are equal: 2 + 7 = 4 + 9. This means that -1 is a good place to start.
2(-1)^3 + 4(-1)^2 + 7(-1) + 5 = -2 + 4 - 7 + 5 = 0. So -1 is a solution. So (x+1) must be a factor. Then you can use synthetic division or polynomial division to get
q(x) = (x+1) (2x^2 + 2x + 5)
A quick check of the discriminant (b^2 - 4ac) for the second factor gives 2^2 - 4.2.5 = -36; since it is negative there are no other real solutions.

For the second question, it looks like you're trying to use the formula (a+b)^n = a^n + ... + (n k)a^k b^(n-k) + ... + b^n, where I'm using (n k) to indicate n choose k. I'm going to assume you know how to evaluate n choose k.
Thus,
(x-1)^6 = x^6 + 6.x^5.(-1) + (6.5/2).x^4.(-1)^2 + (6.5.4 / (3.2.1)).x^3.(-1)^3 + (6.5/2).x^2.(-1)^4 + 6.x.(-1)^5 + (-1)^6
= x^6 - 6 x^5 + 15 x^4 - 20 x^3 + 15 x^2 - 6x + 1.

Answer 3

Sorry, however you've mentioned the situation incorrectly and it are not able to be solved as mentioned. The means it reads proper now's: eighty instances the squareroot of two + 20 the squareroot of ? Please learn situation once more after which put up a correct variation of the situation.

Answer 4

part 2 i guess may be (a+b)^6