Can anybody help me with these math problems? I'm having difficulty with them?

Question

1. Factor this 12cd^3-8c^2 d^2+10c^5 d^3
2. Find the sqaure root of 4a^2+4a+1
3. Multiply (x+3)(X+8)
4. Explain why the product of two conjugates is awlays a rational number
5. Write an equation containing two radicals for where 1 is the solution
6. Find this product (a+b)(a+2b)

Thanks so much for the help guys

Answer 1

1. Factor: 12cd^3 - 8c^2d^2 + 10c^5d^3

First: find the greatest common factor which is, 2cd....

2cd(6d^2 - 4cd + 5c^4d^2)

2. Solve: V`(4a^2+4a+1) > *Factor the expression.

First: multiply the 1st & 3rd coefficient to get "4." Find two numbers that give you "4" when multiplied & "4" (2nd/middle coefficients) when added/subtracted. The numbers are (2 & 2).

Sec: rewrite the expression with the new middle coefficients....

V`4a^2+2a+2a+1

*When you have 4 terms - group "like" terms & factor both sets of parenthesis.

V`(4a^2+2a)+(2a+1)
V`2a(2a+1)+1(2a+1)
V`(2a+1)(2a+1)
V`(2a+1)^2
= 2a+1

3. Multiply (x+3)(x+8)

First: use the foiling method....

(x)(x) + (x)(8) + (3)(x) + (3)(8)
x^2 + 8x + 3x + 24

Sec: combine "like" terms....

x^2 + 11x + 24

6. Find this product (a+b)(a+2b)

First: use the foiling method....

(a)(a) + (a)(2b) + (b)(a) + (b)(2b)
a^2 + 2ab + ab + 2b^2

Sec: combine "like" terms...

a^2 + 3ab + 2b^2

*You need to attempt to solve the rest.

Answer 2

I'll do a few for you...

1)
12cd^3 = 2*2*3*c*d*d*d
8c^2d^2 = 2*2*2*c*c*d*d
10c^5d^3 = 2*5*c*c*c*c*c*d*d*d

each term above has a 2, a "c", and two "d"... So, if you remove that from each one:

12cd^3 = (2cd^2)(2*3*d) = (2cd^2)(6d)
8c^2d^2 = (2cd^2)(2*2*c) = (2cd^2)(4c)
10c^5d^3 = (2cd^2)(5*c*c*c*c*d) = (2cd^2)(5c^4 d)

Put those back together:

(2cd^2)(6d - 4c + 5 c^4 d)

2)
4a^2 + 4a + 1

Factor that:

(2a + 1)(2a + 1)
= (2a + 1)^2

sqrt of that = (2a + 1)

3)
Use FOIL (First, Outside, Inside, Last)

(x + 3)(x + 8)
= x^2 + 8x + 3x + 24
= x^2 + 11x + 24

(do #6 the same way as #3)

Answer 3

I don't think that anyone has answered nos 4 or 5 yet,
so I'll do those.
4.
By "conjugates", I will assume you mean
2 numbers of the form a + b√d and a - b√d,
where a, b and d are rational numbers.
If you multiply these by the foil rule, you get a²-b²d,
which is the difference of 2 rational numbers, so it is rational.

5. How about x + √2 = -1/(x-√2)?
1 is a solution of this equation and so is -1.

Answer 4

gotta go to work, but the first has only some common factors, removed them.

1.They can all give up 2cd^2(6d-4c+5c^4d^3)

the stuff inside the ( ) is not further factorable.

3. x^2+11x+24

6 a^2+3ab+2b^2

5. rad1 times rad 1 = l

gotta go... 2 and 4 would take me more than 3, and I'm late.... hope this helps, hon

Answer 5

i think you're to sparkling up for x? i'm going to do some for you, then you definately would desire to be waiting to be sure something: 3 + 2x sqrt 3 = 5 2x sqrt 3 = 2 x sqrt 3 = a million x = a million/(sqrt 3) 8 + sqrt (x + a million) = 2 sqrt (x + a million) = -6 (sq. the two factors) x + a million = 36 x = 35 sqrt 21 - sqrt (5x - 4) = 0 sqrt 21 = sqrt (5x - 4) (sq. the two factors) 21 = 5x - 4 25 = 5x 5 = x 2sqrt(x + 11) = sqrt (x + 2) + sqrt (3x - 6) (sq. the two factors) 4(x+11) = (x+2) + 2sqrt(x+2)*sqrt(3x-6) + (3x-6) 4x + 40 4 = 4x - 4 + 2sqrt(x+2)*sqrt(3x-6) 40 = 2sqrt(x+2)sqrt(3x-6) (sq. the two factors returned) 1600 = 4(x+2)(3x-6) 4 hundred = (x+2)(3x-6) 4 hundred = 3x^2 - 6x + 6x -12 4 hundred = 3x^2 - 12 412 = 3x^2 412/3 = x^2 x = sqrt (412/3) and -sqrt(412/3)

Answer 6

2. 4a^2+4a+1 = (2a+1)^2 , so + or - (2a+1)
3. (x+3)(x+8) = x^2 + 11x + 24
6. (a+b) (a+2b) = a^2 + 3ab + 2b^2

Answer 7

3. Xx + 8x + 3X + 24 =
unless you meant the x's to be all the same then it would be:

x^ + 8x + 3x + 24 = x^ + 11 x + 24

6. a^ + 2ab + ab + 2b^ = a^ + 3ab + 2b^

Answer 8

3. x^2+11x+24
6. a^2+3ab+2b^2

Answer 9

ans
1)12cd^2(-4c+6d+5c^4d)
2)sqrt(4a^+4a+1)==sqrt((1+2a)^2)== 1+2a
3)xX+8x+3X+24
6)a^2+3ab+2b^2