Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Chapter Review
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Exercise 80 Page 351

Expand the expression by using the Pascal's Triangle and the Binomial Theorem.

108

Practice makes perfect
To find the coefficient of the x^2 term of the binomial expansion, we should recall the Binomial Theorem. It states that for every positive integer n, we can expand the expression (a+b)^n by using the numbers in the n^(th) row of Pascal's Triangle. (a+b)^n = P_0a^nb^0+P_1a^(n-1)b^1+... +P_(n-1)a^1b^(n-1)+P_na^0b^n In the above formula, P_0, P_1, ..., P_n are the numbers in the n^(th) row of Pascal's Triangle.

l&l&l&l&l&l&l&l&l&l&l&l&l&l Row&&&&&&Pascal's&&&&&&& &&&&&&Triangle&&&&&&& c&c&c&c&c&c&c&c&c&c&c&c 0& & & & & &1 & & & & & 1& & & & &1 & &1 & & & & 2& & & &1 & &2 & &1 & & & 3& & & 1 & & 3 & & 3 & & 1 & & 4& &1 & &4 & &6 & &4 & &1 & 5&1 & &5 & &10 & &10 & &5 & &1 Note that each number found in the triangle that is the sum of the two numbers diagonally above it. Now consider the given binomial. ( 3x+ 4 )^3 We can substitute the first term for a and the second term for b using the Binomial Theorem equation and the coefficients from Pascal's Triangle.

(a+b)^n=P_0a^nb^0+P_1a^(n-1)b^1+... +P_(n-1)a^1b^(n-1)+P_na^0b^n
( 3x+ 4)^3= 1( 3x)^3( 4)^0+ 3( 3x)^2( 4)^1+ 3( 3x)^1( 4)^2+ 1( 3x)^0( 4)^3
Finally, let's simplify the expression.
1(3x)^3(4)^0+3(3x)^2(4)^1+3(3x)^1(4)^2+1(3x)^0(4)^3
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Simplify
1(3x)^3(1)+3(3x)^2(4)^1+3(3x)^1(4)^2+1(1)(4)^3
1(3x)^3(1)+3(3x)^2(4)+3(3x)(4)^2+1(1)(4)^3
(3x)^3+3(3x)^2(4)+3(3x)(4)^2+4^3
(3)^3(x)^3+3(3)^2(x)^2(4)+3(3x)(4)^2+4^3
27x^3+3(9)x^2(4)+3(3x)(16)+64
27x^3+108x^2+144x+64
The coefficient of the x^2 term of this expansion is 108.