Algebra Help! (Just explain HOW to do it, not the answer?

Question

I am going into 7th grade and got an algebra packet. I am almost finished, hust need help on some problems. DO NOT state the answer.

(^ means to the power of)

1. 3c^2(4c + 2) + 3c^3 - 3c^2 + 8

2. 3x (x^ + 5) + 7x^3 + 2x -4

3. (5c)(8c^6)

I appreciate ANYONE who helps, no matter how little. Thanks for looking!

correction to #2! it is 3x(x^2+5)......
NOT: 3x(x^+5)...
Sorry!

Answer 1

For all of them you need to distribute most of the values through the () first, so for ex.2 you have 3x(x^5) + 7x^3+2x-4... so you should have 3x^6+7x^3+2x-4...then a common techinque is to break them up with common factors so break it in half with ( ) to get (3x^6 + 7x^3) + (2x-4) like terms in the first bracket you pull out an x^3 and you have x^3(3x^3+7x)+(2x-4). Work from there, for most of them you just need to group them together in order to break them apart easily with common factors of x^? which will always be an x = 0 answer plus the other answers.

Answer 2

1. Remove the parentheses by multiplying 3c^2 by 4c and then adding 3c^2 times 2. Then collect (add) all like terms.

2. Same as problem 1. Just remember that a(b+c) = ab+ac.

3. This is a straight multiplication. Multiply the 5 by the 8 and the c by c^6. Remember that a^m times a^n = a^(m+n).

Answer 3

For the first question, expand the part in brackets by doing 3c^2 * 4c + 3c^2 * 2. In order to multiply with exponents, you have to multiply the big numbers and add the exponents.Then add what you can but remember, you cannot add stuff like 7+4x+4x^2. It's basically the same thing for questions 2 and three. Question two is 3x * x + 3x * 5. (* means multiply) Does that help?

Answer 4

1.3c^2 times 4c then 3c^2 times 2. for the second part,you cant do anything to it so you leave as it is.then you should get something like:12c^3+6c^2+3c^3 - 3c^2 + 8. next you put all the parts with the SAME power together.so you will get: 15c^2+3c^2+8<-----------thats your final answer

2.i think something is missing 3x (x^ + 5) + 7x^3 + 2x -4

3.multiply the coefficient first,so 5 times 8 is 40.then multiply the c's.c times c^6 is c^7.then you add everything together and get the final answer which is 40c^7

Answer 5

You'll need to use the distributive property:
a(b+c) = ab + ac

Examples:

3(2 + 3) = 3*2 + 3*3 = 6 + 9 = 15
2y(3x + 4y^2) = 2y*3x + 2y*4y^2 = 6yx + 8y^3

and use the law of exponents:
a^x * a^y = a^(x+y)

Examples:
2^3 * 2^2 = 2^(3+2) = 2^5
x^3 * x = x^(3+1) = x^4
3y * 2y = 3 * 2 * y^(1+1) = 6 * y^2

Also remember that you can only add like terms that are raised to the same power

Examples:
6x^2 + 10x^2 = 16x^2
3x^3 + 2x^2 + 4x^3 = 7x^3 + 2x^2 (you can not add 7x^3 and 2x^2 b/c they do not have the same power)

Answer 6

alright, you do have to do 3/4-1/2 first. the common denominator is 4, so you change the 1/2 to 2/4 (they're the same thing, but it's easier to subtract 2/4 from 3/4). so, 3/4-2/4--this is easy--is 1/4. after that, you're suppose to multiply the answer by a half. remember "please excuse my dear aunt sally?" (parenthesis, exponents, multiply, divide, add, then subtract) you're suppose to multiply before you add or subtract. so you don't do 2/3-1/2 before you multiply. so, 1/2(1/4) is 1/8 b/c you multiply the numerators together and the denominators together. so, now you do the 2/3-1/8 because it's in parenthesis. the common denominator is 24 because that's the smallest number that both 3 and 8 can evenly divide into. so, you change 2/3 to 16/24 because you have to multiply 3 by 8 to get 24 and you have to multiply the numerator by 8 too to get a fraction that's equal to 2/3. you multiply the top and bottom numbers of 1/8 to get 3/24. Now you have 16/24-3/24 which is 13/24. multiply the top by -3 and the bottom by 5 and get -39/120.

Answer 7

For number 3:

5c = 5c^1 because squaring something to the first degree has no effect. so: (5c^1)(8c^6)

When you multiply somehting like this together, the two basic numbers, 5 & 8, are just multiplied normally. When squares are multiplied, though, you add them together. Say you have the problem: (2x^2) (4x^3) It would equal 8x^5. Double check by replacing x with a number like 2. x here means multiply.(2x2^2)(4x2^3)
(2x4)(4x8)
8x32
256 is your answer. Now for the simplified version we got:
(8x2^5)
8x32
256 you got the same number, so it is correct.
Hope it helped!

Answer 8

when you have a sequence of terms like 3c^2(4c + 2), you want to use what's known as the "distributive property"

lets say i have a(b+c)
i would "distribute" the a to both the b and c
a(b+c) would equal ab+ac
i just multiplied the 'a' by each individual term

lets say i have 5c^2(8c)
when we want to multiply these terms, first i want to multiply the numbers in front of the letters, or constants

5*8 = 40

then i want to multiply the variables
remember, when i multiply two variables of different powers, you just add the exponents

x^n * x^y = x^(n+y)
c^2 * c = c^3

after i multiply the constants and the variables, i simply put them together

5c^2(8c) = 40c^3

Answer 9

1. 12c^3+6c^2+3c^3-3c^2+8=
15c^3+3c^2+8
2. I assume the expression is 3x(x+5)+7x^3+2x-4
and NOT 3x(x^5)+....
3x^2+15x+7x^3+2x-4
=7x^3+3x^2+17x-4
3. 5c(8c^6)=5c*8c^6=40c^7

Answer 10

The first step is to carry out the multiplications and additions.

For instance, if you had (15 b^2)(5b + 3) + 3b^3 + b^2 + 64, you could multiply out the first terms to get 75b^3 + 45b^2, and then add the other terms to get 78b^3 +46 b^2 + 64

Or if you had (9q)(15q^3), you could multiply 9 times 15 to get 135, and multiply q times q^3 to get q^4, and you'd have 135 q^4

Your problems will simplify much more nicely than my examples....

You must be a good student to be getting problems like this in the 7th grade - in my school, these were 9th grade problems.