How do you use the Binomial Theorem to expand # (3x+y)^4#?

1 Answer
Jul 6, 2017

Answer:

# (3x+y)^4 = 81x^4 + 108x^3 y + 54x^2 y^2 + 12x y^3 + y^4 #

Explanation:

Recall the Binomial Theorem

# (a+b)^n = sum_(r=0)^n \ ( (n), (r) ) \ a^(n-r) \ b^r #

Where:

# ( (n), (r) ) =""^nC_r = (n!)/((n-r)!r! #

is the combinatorial, which are also the numbers from the #nth# row of Pascal's Triangle

Thus we have:

# (3x+y)^4 = sum_(r=0)^4 \ ( (4), (r) ) \ (3x)^(4-r) \ (y)^r #

# " " = ( (4), (0) ) (3x)^4 + ( (4), (1) ) (3x)^3 y + ( (4), (2) ) (3x)^2 y^2 + ( (4), (3) ) (3x) y^3 + ( (4), (4) ) y^4 #

# " " = ( 1 ) (3x)^4 + ( 4 ) (3x)^3 y + ( 6 ) (3x)^2 y^2 + ( 4 ) (3x) y^3 + ( 1 ) y^4 #

# " " = ( 1 ) 81x^4 + ( 4 ) 27x^3 y + ( 6 ) 9x^2 y^2 + ( 4 ) 3x y^3 + ( 1 ) y^4 #

# " " = 81x^4 + 108x^3 y + 54x^2 y^2 + 12x y^3 + y^4 #