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 74 Page 351

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

27a^3+27a^2+9a+1

Practice makes perfect

To expand the binomial, 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. ( 3a+ 1 )^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
( 3a+ 1)^3= 1( 3a)^3( 1)^0+ 3( 3a)^2( 1)^1+ 3( 3a)^1( 1)^2+ 1( 3a)^0( 1)^3

Finally, let's simplify the expression.

1(3a)^3(1)^0+3(3a)^2(1)^1+3(3a)^1(1)^2+1(3a)^0(1)^3
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Simplify
1(3a)^3(1)+3(3a)^2(1)^1+3(3a)^1(1)^2+1(1)(1)^3
1(3a)^3(1)+3(3a)^2(1)+3(3a)(1)^2+1(1)(1)^3
(3a)^3+3(3a)^2+3(3a)(1)^2+1^3
(3)^3(a)^3+3(3)^2(a)^2+3(3a)(1)^2+1^3
27a^3+3(9)a^2+3(3a)(1)+1
27a^3+27a^2+9a+1