School and Work

Think of it as being:
x(x^2+x+3) - 3(x^2+x+3)
so then it is more simple

x(x^2+x+3) - 3(x^2+x+3)
= x^3 + x^2 + 3x - 3x^2 -3x - 9

Then group like terms:
x^3 + x^2 + 3x - 3x^2 -3x - 9
= x^3 + (1-3)x^2 + (3-3)x - 9
= x^3 - 2x^2 +0x - 9
= x^3 - 2x^2 - 9

And you're done!

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Home >All Categories > Science & Mathematics >Mathematics >Open Question aba Member since:October 19, 2007Total points:284 (Level 2)9. Let F be a field with p^n elements. Show that F has a subfield K with p^m elements if
and only if m | n.
10. Let K be a finite field. Show that the product of all the nonzero elements of K is -1.
11. Write f(x) = x^16 - x in Z2[x] (that is, f(x) = x^2^4- x) as the product of all monic
irreducible polynomials over Z2 of degree dividing 4.
12. (a) Determine the number of monic irreducible polynomials of degree 6 in Z2[x].
(b) Determine the number of monic irreducible polynomials of degree 11 in Z3[x].Answer Question

Be the first to answer this question.

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6(3x^2 +6) - 4(4x^2 + 5)

expand by distributing the 6 and the - 4 and we get

18x^2 + 36 - 16x^2 - 20

combine like terms (ie the x^2's the x's and the constants) and we get

(18x^2 - 16x^2) + (36 - 20)

2x^2 + 16

or

2(x^2 + 8)

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ok. So if the ^5 is the house and the (x+1) is the man inside the house, you have to take the derivative of the house before the man. This is the chain rule.
using this logic, we get this answer:
5(x+1)^4 * 1
house man

to take the double derivative we use the same process, giving us 20(x+1)^3 *1

the same logic goes for the second one, but instead the chain rule gives us not just a "*1" as it did in the last one. the man inside the house is (x^2-x+1) and the derivative of that is (2x-1) so you have to multiply that one by the exponent derivative, aka the house.
3(x^2-x+1)^2 *(2x-1)
the second derivative of this uses the product rule. f'g+g'f, as well as the chain rule.

You can also do all of this easily on a calculator. especially a TI-89

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