(x^2+x) (x2)
I know that the answer is going to be:x^3x^22x
But im so confused right know! I dont know how to start..

(x^2+x) (x2)
I know that the answer is going to be:x^3x^22x
But im so confused right know! I dont know how to start..
You have to multiply each term in the first bracket by each in the second and vice versa,
So the multiplications you have to do are
x^{2} * x = x^{3}
x * 2 = 2x
x^{2 }*2 = 2x^{2}
x * x = x^{2}
You then add the results of these together: x^{3} + (2x) + (2x^{2})+ x^{2} = x^{3}x^{2}2x
Last edited by PhDemon; February 19th, 2014 at 07:15 AM.
To remember the sequence, I memorised the word FIOB, standing for first, inside, outside, back. Not sure if there was a reason to do it in sequence. Can't think of one actually....
My first algebra teacher used to draw a smiley face with the lines indicating the multiplications to do, lines above the brackets going from the x^2 to the x (in second bracket) and the x(in first bracket) to the 2 as the eyebrows. Below you draw two lines, one between the x's and one from the x^2 to the 2, this is the smile
The order you do the steps in doesn't matter (as long as they are all included) as you add all the results up at the end.
FOIL (first, outer, inner, last) might be an easier word to remember if you want a word mnemonic.
Last edited by MagiMaster; February 19th, 2014 at 10:09 AM. Reason: Typo
x^3  2x^2 + x^2  2x
but if the superscript are the same, cant i just add x^3+x^2 so i get x^5?
No, you only add the exponents (superscripts) if you are multiplying. If they are the same you add the numbers in front of the x^n to get the answer, in the example you have given the superscripts are NOT the same (they are 3 and 2).
Examples:
x^3 * x^2 = x^5
x^3 + x^2 = x^3 + x^2 (not x^5)
2x^2 + x^2 = 3x^2
etc...
sunshine345, It might help if you expand the exponents. x^2 is the same thing as x*x. x^3 is x*x*x. So like PhDemon mentioned:
x^3 * x^2 = x*x*x * x*x = x^5
x^3 + x^2 = x*x*x + x*x, which you can't exactly simplify further.
If you think about it, everyone learns the basic concept in about Grade 2 or 3 (the teachers simply didn't tell us so).
345 × 8006 = (300+40+5) × (8000+6) = (3∙10^{2}+4∙10+5) × (8∙10^{3}+6) =
= (3∙10^{2}×8∙10^{3} + 3∙10^{2}×6) + (4∙10×8∙10^{3} + 4∙10×6) + (5×8∙10^{3} + 5×6) =
= 24∙10^{5} + 18∙10^{2} + 32∙10^{4} + 24∙10 + 40∙10^{3} + 30 =
= 24∙10^{5} + 32∙10^{4} + 40∙10^{3} + 18∙10^{2} + 24∙10 + 30 =
= 2,762,070
If algebra teachers simply revisited this kind of grammar school problem in this way, algebra students would grasp the algebraic concept much more easily.
So let's do this properly.
There exists a very important theorem in arithmetic  so important it's called the Fundamental Theorem of Arithmetic  that states that any Real number can be factored of a product of positive primes together with a unit (). Where the number in question is itself prime, then the the factors are obviously itself and .
A similar rule applies for polynomials. Consider the polynomial given in the OP : . As this can be factored as one says that this polynomial is reducible. Note that the question as to whether an arbitrary polynomial is reducible or not is somewhat vexed (Eisenstein gives a clumsy algorithm  no that's not Einstein)
Note that the factors in the OP ( and ) are themselves polynomials and irreducible. This applies generally, granted the vexation noted above.
In other words, if a polynomial is irreducible we may think of it as being in some sense a "prime polynomial"
Yes, (tautologically) true. I specified the Real numbers, but I misspoke  I really meant the Fundamental Thm usually applies to the Integers.
But look  I neglected to mention the most important part of this Theorem. Which is that....
The factorization of any integer as a product of primes is unique (provided we don't care in which order the factors are written). The proof is straightforward  use the cancellation law for integral domains to find a contradiction when you assume the contrary
This fact BTW is extremely important in cryptography
However, it is not obvious that the factorization of polynomials as the product of irreducible polynomials is unique. In fact, since the integerrs are ring, it might appear that the factorization of any polynomial with coefficients drawn from any ring is unique, just like the above. This is false, apparently
However, it can be shown that when the coefficients are drawn from a field the uniqueness follows  I have a proof here (courtesy Ian Stewart) if you're interested, but it's involved
Solve for x: x2 = 2x (x square = 2x)
you can solve this 3 ways, which one is right and why.
1) Take the square root of both sides, which give you x= (sq root) of 2x
2) Divide both sides by x, which gives x = 2
3) or move the 2x over and factor. (x0)(x2)
Secondly, when you factor this problem, how come the sum of the terms doesn't add up to 2x as the FOIL method says.
The first way still leaves you with x = +sqrt(2x) or x = sqrt(2x) so you haven't completely isolated x.
The second way is fine except you are assuming x is not zero or you couldn't divide both sides by it.
The third way is correct and gives you the solution x = 0 and x = 2.
Using FOIL on (x0)*(x2) = 0 you have x*x  x*2  0*x + (0*2) = 0.
x^2  2x  0 + 0 = 0
x^2  2x = 0
x2 = 2x
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