1. (x^2+x) (x-2)

I know that the answer is going to be:x^3-x^2-2x

But im so confused right know! I dont know how to start..  2. ### Related Discussions:

3. 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

x2 * x = x3

x * -2 = -2x

x2 *-2 = -2x2

x * x = x2

You then add the results of these together: x3 + (-2x) + (-2x2)+ x2 = x3-x2-2x  4. 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....  5. 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   6. Originally Posted by PhDemon 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 Yeah that's the one. Any reason to do it in sequence? First the first two terms, the inside ones, then...etc?  7. 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.  8. Originally Posted by PhDemon 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.
Thought so, thanks.  9. FOIL (first, outer, inner, last) might be an easier word to remember if you want a word mnemonic.  10. Originally Posted by MagiMaster FOIL (first, outer, inner, last) might me an easier word to remember if you want a word mnemonic.
True. I had to translate it from Afrikaans, which was EBBA.  11. 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?  12. 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...  13. Originally Posted by sunshine345 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?
Work it out and see. Substitute x with 2 and work both cases out. The answers don't match (x3+x2 is 12, while x5 is 32), so the answer is no. This is basic exponential rules though.  14. 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.  15. Thanks! I get it now   16. 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∙102+4∙10+5) × (8∙103+6) =

= (3∙102×8∙103 + 3∙102×6) + (4∙10×8∙103 + 4∙10×6) + (5×8∙103 + 5×6) =

= 24∙105 + 18∙102 + 32∙104 + 24∙10 + 40∙103 + 30 =

= 24∙105 + 32∙104 + 40∙103 + 18∙102 + 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.  17. 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"  18. Originally Posted by Guitarist In other words, if a polynomial is irreducible we may think of it as being in some sense a "prime polynomial"
Just be careful, being prime is not the same as being irreducible - in an integral domain every prime is irreducible but you need a unique factorization domain for the inverse to be true.  19. Originally Posted by river_rat in an integral domain every prime is irreducible but you need a unique factorization domain for the inverse to be true.
Yes, (tautologically) true. I specified the Real numbers, but I mis-spoke - 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  20. 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. (x-0)(x-2)
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 (x-0)*(x-2) = 0 you have x*x - x*2 - 0*x + (0*2) = 0.
x^2 - 2x - 0 + 0 = 0
x^2 - 2x = 0
x2 = 2x  Bookmarks
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