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# I have a few ASAP questions!!!! Can I get some help?

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What is the coefficient of xy in the expansion of  \((3x+(2y+1))^2\)?

Dec 30, 2017

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Here is a long and detailed answer. Sorry, no LaTex!.

Expand the following:

(3 x + (2 y + 1))^2

Treating (1 + 3 x + 2 y)^2 as the binomial ((3 x + 1) + 2 y)^2, expand using FOIL.

((3 x + 1) + 2 y) ((3 x + 1) + 2 y) = (3 x + 1) (3 x + 1) + (3 x + 1) (2 y) + (2 y) (3 x + 1) + (2 y) (2 y) = (3 x + 1)^2 + 2 (3 x + 1) y + 2 (3 x + 1) y + 4 y^2 = (3 x + 1)^2 + 4 (3 x + 1) y + 4 y^2:

2×2 y (3 x + 1) + (3 x + 1)^2 + (2 y)^2

Multiply 3 x + 1 and 3 x + 1 together using FOIL.

(3 x + 1) (3 x + 1) = (3 x) (3 x) + (3 x) (1) + (1) (3 x) + (1) (1):

2×2 y (3 x + 1) + 3×3 x x + 3 x + 3 x + 1 + (2 y)^2

Combine products of like terms.

3 x×3 x = 3×3 x^2:

2×2 y (3 x + 1) + 3×3 x^2 + 3 x + 3 x + 1 + (2 y)^2

2×2 y (3 x + 1) + 9 x^2 + 3 x + 3 x + 1 + (2 y)^2

Distribute exponents over products in (2 y)^2.

Multiply each exponent in 2 y by 2:

2×2 y (3 x + 1) + 9 x^2 + 3 x + 3 x + 1 + 2^2 y^2

2×2 y (3 x + 1) + 9 x^2 + 3 x + 3 x + 1 + 4 y^2

Group like terms in 2 (3 x + 1)×2 y + 9 x^2 + 3 x + 3 x + 1 + 4 y^2.

Grouping like terms, 2 (3 x + 1)×2 y + 9 x^2 + 3 x + 3 x + 1 + 4 y^2 = 4 y^2 + 4 (3 x + 1) y + 9 x^2 + (3 x + 3 x) + 1:

4 y^2 + 4 y (3 x + 1) + 9 x^2 + (3 x + 3 x) + 1

Add like terms in 3 x + 3 x.

3 x + 3 x = 6 x:

4 y^2 + 4 y (3 x + 1) + 9 x^2 + 6 x + 1

Distribute 4 y over 3 x + 1.

4 y (3 x + 1) = 4 y×1 + 4 y×3 x:

4 y^2 + 4 y + 4×3 y x + 9 x^2 + 6 x + 1

4 y^2 + 4 y + 12 y x + 9 x^2 + 6 x + 1

Dec 30, 2017
#2
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4 y^2 + 4 y + 12 y x + 9 x^2 + 6 x + 1

Is not right, any other ideas???

Dec 31, 2017
#3
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Note  ...  (a  + b)^2  =  a^2  + 2ab  +  b^2

Let  a  =  3x      and b  =  (2y + 1)     ....   so we have

(3x  + (2y + 1) )^2  =

(3x)^2   +  2 (3x)(2y+1)  + (2y + 1)^2

We only  care about the middle term....so....

2(3x)(2y + 1)   =

6x (2y + 1)  =

12xy  +  6x

So.....the coefficient is  12   Dec 31, 2017