Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
Chapter Review
Continue to next subchapter

Exercise 81 Page 351

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

6a^2c^2

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. ( ax -c )^4 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
( ax -c)^4= 1( ax)^4( - c)^0+ 4( ax)^3( - c)^1+ 6( ax)^2( - c)^2+ 4( ax)^1( - c)^3+ 1( ax)^0( - c)^4

Finally, let's simplify the expression.

1(ax)^4(- c)^0+4(ax)^3(- c)^1+6(ax)^2(- c)^2+4(ax)^1(- c)^3+1(ax)^0(- c)^4
â–¼
Simplify
1(ax)^4(1)+4(ax)^3(- c)^1+6(ax)^2(- c)^2+4(ax)^1(- c)^3+1(1)(- c)^4
1(ax)^4(1)+4(ax)^3(- c)+6(ax)^2(- c)^2+4(ax)(- c)^3+1(1)(- c)^4
(ax)^4+4(ax)^3(- c)+6(ax)^2(- c)^2+4(ax)(- c)^3+(- c)^4
(a)^4(x)^4+4(a)^3(x)^3(- c)+6(a)^2(x)^2(- c)^2+4(ax)(- c)^3+(- c)^4
a^4x^4+4a^3x^3(- c)+6a^2x^2c^2+4ax(- c^3 )+c^4
a^4x^4-4a^3x^3c+6a^2x^2c^2-4axc^3+c^4
a^4x^4-4a^3cx^3+6a^2c^2x^2-4ac^3x+c^4

The coefficient of the x^2 term of this expansion is 6a^2c^2.