Can somebody help me solving this question

Find the term in x^3 in the expansion of (1+5x)^2(1-2x)^6

For a binomial expansion (a + b)^n:

-- There are n + 1 terms
-- The sum of the exponents in each term is n
-- The exponent of the "b" term is one less than the number of the term, e.g., for the 1st term, the exponent of b is 0; for the 2nd term, the exponent of b is 1, etc.
-- The coefficient in front of the term is (n) C (either exponent value), e.g., for a^3b^2, the exponent is 5C3 = 10 or 5C2 = 10.


Let's first look at (1 - 2x)^6:

[(1) + (-2x)]^6.....where a = 1, b = -2x, and n = 6


-- There are 7 terms:

6C0 (1)^0(-2x)^6 + 6C1 (1)^1(-2x)^5 + 6C2 (1)^2(-2x)^4 + 6C3 (1)^3(-2x)^3 + 6C4 (1)^4(-2x)^2 +
6C5 (1)^5(-2x)^1 + 6C6 (1)^6(-2x)^0

= 1 (-2x)^6 + 6 (-2x)^5 + 15 (-2x)^4 + 20 (-2x)^3 + 15 (-2x)^2 + 6 (-2x)^1 + 1

= 64x^6 - 192x^5 + 240x^4 -160x^3 + 60x^2 - 12x + 1


Now let's look at (1 + 5x)^2.....[multiplying like the first polynomial and not using (1 + 5x)(1 + 5x)]:

2C0 (1)^0(5x)^2 + 2C1 (1)^1(5x)^1 + 2C2 (1)^2(5x)^0

= (5x)^2 + 2 (5x) + 1

= 25x^2 + 10x + 1


So, the x^3 term in the expansion will involve the following terms:

(25x^2)(-12x) + (10x)(60x^2) + (1)(-160x^3)

= -300 x^3 + 600 x^3 - 160 x^3

= 140 x^3


Therefore, the x^3 term in the expansion is 140 x^3.



Hope this helps...good luck!