# Factor completely 18a^2+60ab+50b^2?

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Given the expression 18a^2+60ab+50b^2:

Method I: Factor out the greatest common factor; in this case 2, thus 2(9a^2+30ab+25b^2). We then note that the expression in parantheses is a perfect square trinomial of the form x^2+2xy+y^2 which factors as (x+y)^2. Thus letting x=3a and y=5b we get a result of 2(3a+5b)^2.

Method II: Multiply 18*50=900. We look for numbers whose product is 900 and whose sum is 60; the numbers are 30 and 30. We rewrite the expression as 18a^2+30ab+30ab+50b^2 and factor pairwise getting 6a(3a+5b)+10b(3a+5b). Noting that the expressions in the parantheses are the same we use the distributive property to get (3a+5b)(6a+10b). Factoring a 2 from the second factor yields 2(3a+5b)(3a+5b)=2(3a+5b)^2.

First, we'll factor the given expression by 2:

`18a^2 + 60ab + 50b^2 =` 2(` ` `9a^2 + 30ab + 25b^2` )

We notice that we've get within brackets a perfect square: `(a+b)^2 = a^2 + 2ab + b^2`` `** **`2(3a + 5b)^2 = 2(9a^2 + 30ab + 25b^2)`

**Therefore, the completely factorised expression is: `18a^2 + 60ab + 50b^2 = 2(3a + 5b)^2` **

I like the factoring by grouping method:

Multiply the first coeffecient, 18, by the last, 50 and we get 900.

We to find two numbers that MULTIPLY to equal 900 and ADD to equal 60 (the middle term)

We can use 30 and 30 because 30*30= 900 and 30+30 =60

Now we will rewrite 60ab as 30ab+30a giving us

###### 18a^2 + 30ab + 30ab +50b^2

Group the first two and last two terms such as

(18a^2 + 30ab) + (30ab +50b^2)

Factor each set of () to get

{6a(3a+5b)} +{10(3a+5b)}

Notice that we have a (3a+5b in each set of {}, so we can factor that out and get

{6a+10b}(3a+5b)

we can continue to factor {6a+10b} as

2{3a+5b}(3a+5b)

Now we can rewrite {3a+5b}(3a+5b) using an exponent to get our final answer of

2(3a+5b) ^ 2