simplify (2n 3 – 5)(6n 2 + n), 6k(4k + k 2) + 9k(2k – 6k 2),
- Thread starter eboo1786
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arthur ohlsten
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Re: PLEASE HELP 
I can only simplify the last 2.
6k[4k+k^2]+9k[2k-6k^2]
factor k from first bracket and 2k from second
6k^2[4+k]+18k^2[1-3k]
factor out 6k^2
6k^2{ [4+k] +3[[1-3k]}
remove inner brackets
6k^2{4+k+3-9k}
6k^2[7-8k] answer
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[4x^2+x+3]+[5x^2+9x-2] remove brackets
4x^2+x+3+5x^2+9x-2 combine likes
9x^2+10x+1 factor
factors of 9=1,9.... 3,3
factors of 1=1,1
we want a set whose sum is 10
9,1 and 1,1 does it
9x^2+10x+1=[9x+1][x+1] answer
I can only simplify the last 2.
6k[4k+k^2]+9k[2k-6k^2]
factor k from first bracket and 2k from second
6k^2[4+k]+18k^2[1-3k]
factor out 6k^2
6k^2{ [4+k] +3[[1-3k]}
remove inner brackets
6k^2{4+k+3-9k}
6k^2[7-8k] answer
_______________________________________________________
[4x^2+x+3]+[5x^2+9x-2] remove brackets
4x^2+x+3+5x^2+9x-2 combine likes
9x^2+10x+1 factor
factors of 9=1,9.... 3,3
factors of 1=1,1
we want a set whose sum is 10
9,1 and 1,1 does it
9x^2+10x+1=[9x+1][x+1] answer
Re: PLEASE HELP
Ok, I will help you with the first one. If you need to simplify a product of two or more sets of parantheses, first of all you need to get rid of those brackets. To do that, you must do some calculus, which is: you multiple each member from each paranthese with all the members from the other(s) paranthese(s), then you add this product. Of course, each time you multiply max two products at once.
Example: (a-b)(c+d) = ac + ad -bc - bd. (1) (I multiplied a with c, a with d, -b with c, -b with d, and then I added these sums) Another example: (a-b)(c+d)(e+f+g). First we do (a-b)(c+d), then the result is multiplied with (e+f+g). From (1) => (ac + ad - bc - bd) (e + f + g). What was the rule? Each member from the first paranthese is multiplied with each member from the second one, and those results are added:
ace + acf + acg + ade + adf + adg - bce - bcf - bcg - bde - bdf - bdg. (2)
Hope you understood so far. A, when you do these multiplications, you take into consideration the numbers WITH their signs. For example, for the bc term, I took -bc. That' s why in my addition from (2) I got also substraction. But the substraction is the addition with the changed sign: a-b = a + (-b).
So referring to your example, (2n 3 – 5)(6n 2 + n) we do the same thing. We do the multiplication between each term from the first set of brackets, with each term from the second one, and then, of course, we add those sums: We multiply 2n^3 with 6n^2, 2n^3 with n, -5 with 6n^2 and -5 with n. Then we get 12n^5 + 2n^4 - 30n^2 - 5n. That' s all.
eboo1786 said:
Ok, I will help you with the first one. If you need to simplify a product of two or more sets of parantheses, first of all you need to get rid of those brackets. To do that, you must do some calculus, which is: you multiple each member from each paranthese with all the members from the other(s) paranthese(s), then you add this product. Of course, each time you multiply max two products at once.
Example: (a-b)(c+d) = ac + ad -bc - bd. (1) (I multiplied a with c, a with d, -b with c, -b with d, and then I added these sums) Another example: (a-b)(c+d)(e+f+g). First we do (a-b)(c+d), then the result is multiplied with (e+f+g). From (1) => (ac + ad - bc - bd) (e + f + g). What was the rule? Each member from the first paranthese is multiplied with each member from the second one, and those results are added:
ace + acf + acg + ade + adf + adg - bce - bcf - bcg - bde - bdf - bdg. (2)
Hope you understood so far.
A, when you do these multiplications, you take into consideration the numbers WITH their signs. For example, for the bc term, I took -bc. That' s why in my addition from (2) I got also substraction. But the substraction is the addition with the changed sign: a-b = a + (-b). So referring to your example, (2n 3 – 5)(6n 2 + n) we do the same thing. We do the multiplication between each term from the first set of brackets, with each term from the second one, and then, of course, we add those sums: We multiply 2n^3 with 6n^2, 2n^3 with n, -5 with 6n^2 and -5 with n. Then we get 12n^5 + 2n^4 - 30n^2 - 5n. That' s all.