Find the factors of the following: c2 + 2cd +d2=?

Question

Help?! I don't get this, this is my second sheet and I don't get this one. It's nothing like the 1st one I got. You guys helped me with it earlier today. The problems on this is x2 + 4xy +4y2 =?

The 2 are squared and the c2 in the 1st problem is squared too, can someone help me understand how to find the factors? Like what? I'm not getting this at all. I just need to see it step by step and explain and then I will be able to do the 24 other ones. Thanks you so much!

Answer 1

c^2 + 2cd + d^2 =
(c + d)(c + d)

Answer 2

c^2 + 2cd + d^2 has as factors (c+d)(c+d), usually written as (c+d)^2. Now to answer your question.

An algebraic expression has as many factors as the highest power found in the expression.In both your examples the highest power is ^2. Therefore both expressions will have 2 factors.Let's write these two examples down again side by side
c^2 +2cd +d^2 x^2 +4xy + y^2
( )( ) ( )( )
The trick is to write in just the letters that will give you the first and last terms of your expression,like so
(c )(c ) (x )(x )
(c d)(c d) (x y)(x y)

Notice I've avoided including any signs in my work so far. That comes very soon.
NOW I'm assuming you know how to multiply two binomials together. Some teachers call it the FOIL method. First terms, Outside terms, Inside terms,
Last terms. FOIL
Example:
(X+2y)(x-5y) is Fx.x, O-5xy, I 2xy, L -10y^2
(x + 2y)(x-5y) = x^2 -5xy+2xy -10y^2
(x + 2y)(x-5y) = x^2 -3xy -10y^2

Suppose x^2 -3xy -10y^2 was the expression you had to factor-how would you do it?Obviously we have to end up with (x+2y)(x-5y). We do it this way:
x^2 -3xy -10y^2
(x y)(x y)
Here's the key move.
"We need two numbers that multiply to give -10 and add to give -3. With a bit of thought, we come up with 5 and 2. We still need the correct signs to go in front of each number, so we have a bit more thinking to do. Well to get a - answer for -10, either the 5 has to be - or the 2 has to be -. Now since they have to add up to -3, it requires that the 5 be minus. The 2 has to be +
So we get (x+2)(x-5).

Now back to your examples
c^2 +2cd + d^2 = (c d)(c d). We need two numbers that multiply to give +1, and add up to +2. The numbers are 1 and 1
(c+d)(c+d)

x^2 + 4xy + 4y^2 = (x y)(x y)
We need two numbers that multiply to give +4 and addup to +4.
The numbers are 2 and 2
(x + 2y)(x + 2y)

In both of your examples the 2 factors are the same.
That's O.K. That's allowed. In fact that"s what you always get when you factor a perfect square. Which your 2 examples are.

Hope this is of use to you. Good luck!

Answer 3

a million. c because of the fact -(a-b) = -a +b = b-a 2. while you're doing away with the undemanding section (p+3q), you're left with a million(from the 1st section) and -r (from the 2d) so (p+3q)(a million-r) 3. (a2 - 4)(a - 2) (a2 + 4)(a - 2) (a2 + 4)(a + 2) i'm assuming those are the three diverse answer possibilities. it must be the middle one because of the fact the final portion of the question is -8. you will in basic terms get that via multiplying the -2 and +4. 4. if the two factors are meant to be squared, then: that's a. everytime you have something in the form(x^2 -y^2) (the place ^2 potential that it truly is squared), you additionally can write it like: (x+y) (x-y). on your problem in basic terms say that x = b and y = c-2d.

Answer 4

if you can take the square root of the first and last terms and the middle term is the addition of the two roots, you have a perfect square trinomial.

c^2+2cd+d^2 = (c+d)(c+d)

x^2+4xy+4y^2 = (x+2y)(x+2y)

to check your answers you can always multiply the brackets (expand) and you should end up with the original trinomial.

Answer 5

c^2 + 2cd + d^2
You obviously need two factors starting with c to get the c^2:
(c + ?)(c + ?)
Similarly, to get d^2, you need a d in each bracket:
(c + d)(c + d)
Multiplying and adding the inside and outside pairs gives you 2cd, and fortunately that is what you want.
Just tidy up the result a bit, as the two factors are the same, and write it:
(c + d)^2.

Answer 6

(c+d)(c+d) = c^2 + 2cd +d^2 = (c + d) ^2
(x + 2y)(x+2y) = x^2+ 4xy + 4y^2 = (x+y)^2

binomial thoerum.

Answer 7

c^2+2cd+d^2
Basically this is a formula
(a+b)^2=a^2+2ab+b^2
so c^2+2cd+d^2
=c^2+cd+cd+d^2
=C(c+d)+d(c+d)
=(c+d)(c+d)
so solving x^2+4xy+4Y^2,we get
=x^2+2xy+2xy+4y^2
=x(x+2y)+2y(x+2y)
=(x+2y)(x+2y) Ans

Answer 8

we have the formula is: a^2+2ab+b^2=(a+b)^2
in this problem:
c^2+2cd+d^2
a in this one is c, b is d
follow the formula, we'll have:
c^2+2cd+d^2= (c+d)^2

Answer 9

(c+d) * (c+d)

c times c = c2
c times d = cd
and d times c = cd
d times d = d2
add all together c2 +2cd + d2

Answer 10

Breaking it down from the answer:

c^2 + 2cd + d^2
c^2 +cd +cd + d^2
(cxc)+ (cxd) + (cxd) + (dxd)
(c+d)(c+d)