SOLUTION: I was instructed to "multiply as required and collect terms" I am really not sure what it means to collect terms. (x–4)*(2x^2+4)+(x+4)*(x^3–1)–(x^2–x–1)

Algebra ->  Exponents -> SOLUTION: I was instructed to "multiply as required and collect terms" I am really not sure what it means to collect terms. (x–4)*(2x^2+4)+(x+4)*(x^3–1)–(x^2–x–1)      Log On


   



Question 187560: I was instructed to "multiply as required and collect terms" I am really not sure what it means to collect terms.
(x–4)*(2x^2+4)+(x+4)*(x^3–1)–(x^2–x–1)

Found 2 solutions by vleith, solver91311:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
I assume you understand the 'multiply' this part. Basically use the same rules you know from arithmetic (distribution, rules of order, etc) to expand all the terms.
Collecting terms means to group like terms and simplify. SO you use assoications and commutation to 'gather up' all the constants (numbers), all the terms with just a single x in them, all the terms with x^2, etc. Once you group them like that, then just add/subtract to simplify
[(x–4)*(2x^2+4)] + [(x+4)*(x^3–1)] – (x^2–x–1) expand (added [] for effect)
2x%5E3+%2B4x+-+8x%5E2+-+16+%2B+x%5E4-x+%2B+4x%5E3+-1+-+x%5E2+%2Bx+%2B+1 group
x%5E4+%2B+2x%5E3+%2B+4x%5E3+-+8x%5E2+-+x%5E2+%2B+4x+-x+%2Bx+-16+-1+-1 simplify
x%5E4+%2B+6x%5E3+-9x%5E2+%2B4x+-18

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




Use FOIL to multiply your binomials:




Now you can remove the parenthesese because you no longer have to segregate things:



"Like Terms" are terms that have the same exponent on the variable. Start with the highest exponent. You have 1 term:



And 2 terms containing , namely and which added together make



And 2 terms containing , namely and which added together make



And 3 terms containing , namely , , and which added together make



And finally you have 3 terms that contain no variable at all (called the constant terms), namely -16, -4, and - 1 which added together make -21.



Technically speaking, the constant terms actually do have the variable in them. It is just that the variable is and anything raised to the zero power is 1, so we just leave it out.

John