Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
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Exercise 77 Page 351

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

243a^5+1620a^4b+4320a^3b^2+5760a^2b^3+3840ab^4+1024b^5

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+ 4b )^5 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+ 4b)^5= 1( 3a)^5( 4b)^0+ 5( 3a)^4( 4b)^1+ 10( 3a)^3( 4b)^2+ 10( 3a)^2( 4b)^3+ 5( 3a)^1( 4b)^4+ 1( 3a)^0( 4b)^5

Finally, let's simplify the expression.

1(3a)^5(4b)^0+5(3a)^4(4b)^1+10(3a)^3(4b)^2+10(3a)^2(4b)^3+5(3a)^1(4b)^4+1(3a)^0(4b)^5
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Simplify
1(3a)^5(1)+5(3a)^4(4b)^1+10(3a)^3(4b)^2+10(3a)^2(4b)^3+5(3a)^1(4b)^4+1(1)(4b)^5
1(3a)^5(1)+5(3a)^4(4b)+10(3a)^3(4b)^2+10(3a)^2(4b)^3+5(3a)(4b)^4+1(1)(4b)^5
(3a)^5+5(3a)^4(4b)+10(3a)^3(4b)^2+10(3a)^2(4b)^3+5(3a)(4b)^4+(4b)^5
(3)^5(a)^5+5(3)^4(a)^4(4b)+10(3)^3(a)^3(4)^2(b)^2+10(3)^2(a)^2(4)^3(b)^3+5(3a)(4)^4(b)^4+(4)^5(b)^5
243a^5+5(81)a^4(4b)+10(27)a^3(16)b^2+10(9)a^2(64)b^3+5(3a)(256)b^4+1024b^5
243a^5+1620a^4b+4320a^3b^2+5760a^2b^3+3840ab^4+1024b^5