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 72 Page 351

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

x^3+27x^2+243x+729

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. ( x+ 9 )^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
( x+ 9)^3= 1 x^3( 9)^0+ 3 x^2( 9)^1+ 3 x^1( 9)^2+ 1 x^0( 9)^3
Finally, let's simplify the expression.
1x^3(9)^0+3x^2(9)^1+3x^1(9)^2+1x^0(9)^3
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Simplify
1x^3(1)+3x^2(9)^1+3x^1(9)^2+1(1)(9)^3
1x^3(1)+3x^2(9)+3x(9)^2+1(1)(9)^3
x^3+3x^2(9)+3x(9)^2+9^3
x^3+3x^2(9)+3x(81)+729
x^3+27x^2+243x+729