How do you factor these three problems....?

Question

1.) x^2y-yz-xyz-x^2y -------I have been trying to figure out this problem but cant get the - + signs right at the end.

2.) x^3+x^2+x+1 ------cant figure it out, would you do this by grouping..?

3.) 12ac+6ad+4bd+bc


Thank you very much!

Answer 1

1) did you miss something on the first 1, since the first and last can be cancel out
then you factor
-zy(1+x)

2) yes you group it first
(x^3+x^2)+(x+1)
x^2(x+1) + 1(x+1)
now factor (x+1)
you get
(x+1)(x^2+1)

3) don't see how you can factor it...the coefficients don't seem to have patterns and 4 variables.

Answer 2

PROBLEM 1:

Do you really mean:
x^2y - yz - xyz - x^2y

If so, the first and last terms cancel:
-yz -xyz

Now pull out a common -yz as a factor:
-yz(1) - yz(x)

Here's the final answer:
-yz(1 + x)

PROBLEM 2:

Factor the first two terms by pulling out a common x^2:
x^2(x + 1) + x + 1

Now notice you have x + 1 repeated, so factor that out:
(x^2 + 1)(x + 1)

PROBLEM 3:

12ac + 6ad + 4bd + bc

Hmm... this ones tougher... are you sure you typed it correctly. I tried factoring out 6a from the first two terms:
6a(2c + d) + 4bd + bc.

Then trying to pull a b from the last two terms:
6a(2c + d) + b(4d + c)

But then I don't see how you can combine anything...

Answer 3

1.) x^2y-yz-xyz-x^2y
First notice that the first and last terms cancel each other out.
So you are left with -yz-xyz = -yz(1 + x)

2.) x^3+x^2+x+1
= x^2(x+1) + 1*(x+1)
=(x^2+1)(x+1)

3.) 12ac+6ad+4bd+bc
= 6a(2c+d) + b(4+c)
Are you sure you copied this last problem correctly?

Answer 4

1) x^2y-yz-xyz-x^2y

First & last terms cancel - did you state this correctly?


2) x^3+x^2+x+1

x^2(x + 1) + 1(x + 1)
(x^2 + 1)(x + 1)


3) 12ac+6ad+4bd+bc

Are you sure you stated this one correctly?

Answer 5

#1
ok u need to clarify on #1 if its x^2y-yz-xyz-x^2y or (x^2)y-yz-xyz-(x^2)y

#2 u do it in groups of 2 so
(x^3+x^2)+(x+1)
x^2(x+1)+1(x+1)
(x^2+1)(x+1)

#3
i have no idea
sry wasn't much help

Answer 6

1) the first and the last term cancel out so you are left with
-yz - xyz -- factor out a yz

(yz) * (-1-x)

2) dont think you can factor it

3) factor by grouping

(6a*(2c+d)) +(b*(4d+c))

not sure if you can go further than that

Answer 7

a million. u(v-a million) - 2(a million-v) = u(v-a million) + 2(v-a million) [each and every time era has (v-a million) factor] = (v-a million)(u+2) 2. z^2 + 2z + a million - w^2 = (z+a million)^2 - w^2 [difference of squares] = [(z+a million) - w][(z+a million)+w] 3. u^2 - 10u +9 = (u-a million)(u-9) verify is left for you.