Asked by Tamara GreenE on Sep 23, 2024
Verified
Use Pascal s Triangle to expand the expression (2x−4) 5( 2 x - 4 ) ^ { 5 }(2x−4) 5 .
A) 2x5−20x4+160x3−640x2+1280x−10242 x ^ { 5 } - 20 x ^ { 4 } + 160 x ^ { 3 } - 640 x ^ { 2 } + 1280 x - 10242x5−20x4+160x3−640x2+1280x−1024
B) 2x5−5x4+10x3−10x2+5x−−42 x ^ { 5 } - 5 x ^ { 4 } + 10 x ^ { 3 } - 10 x ^ { 2 } + 5 x - - 42x5−5x4+10x3−10x2+5x−−4
C) 32x5−80x4+80x3−40x2+10x−102432 x ^ { 5 } - 80 x ^ { 4 } + 80 x ^ { 3 } - 40 x ^ { 2 } + 10 x - 102432x5−80x4+80x3−40x2+10x−1024
D) 32x5−102432 x ^ { 5 } - 102432x5−1024
E) 32x5−320x4+1280x3−2560x2+2560x−102432 x ^ { 5 } - 320 x ^ { 4 } + 1280 x ^ { 3 } - 2560 x ^ { 2 } + 2560 x - 102432x5−320x4+1280x3−2560x2+2560x−1024
Pascal's Triangle
A triangular array of numbers where each number is the sum of the two numbers directly above it, often used in probability theory and algebra.
Expand
To expand an algebraic expression means to simplify the expression by distributing multiplication over addition or subtraction and combining like terms.
- Implement the principles of the binomial theorem and Pascal's triangle to address challenges in binomial expansion.
Verified Answer
MG
Mohit Goyal2 days ago
Final Answer :
E
Explanation :
To expand (2x−4)5(2x-4)^5(2x−4)5 using Pascal's triangle, we start by writing the first six rows of the triangle:
11112113311464115101051\begin{matrix}& & & & 1 & & & & \\& & & 1 & & 1 & & & \\& & 1 & & 2 & & 1 & & \\& & 1 & & 3 & & 3 & & 1 \\& & 1& & 4 & & 6 & & 4 & & 1 \\& & 1 & & 5 & & 10 & & 10 & & 5 & & 1 \\\end{matrix}11111123451136101410151
The coefficients of the expanded expression will be the entries in the 6th row of Pascal's triangle: 1, 5, 10, 10, 5, 1. We will also have powers of 2x2x2x and powers of −4-4−4 , starting with 5 and decreasing to 0.
Using the row of coefficients and powers, we can build the expanded expression term by term. The first term is:
(2x)5=32x5(2x)^5 = 32x^5(2x)5=32x5
The second term is:
(51)(2x)4(−4)1=−320x4\binom{5}{1}(2x)^4(-4)^1 = -320x^4(15)(2x)4(−4)1=−320x4
The third term is:
(52)(2x)3(−4)2=1280x3\binom{5}{2}(2x)^3(-4)^2 = 1280x^3(25)(2x)3(−4)2=1280x3
The fourth term is:
(53)(2x)2(−4)3=−2560x2\binom{5}{3}(2x)^2(-4)^3 = -2560x^2(35)(2x)2(−4)3=−2560x2
The fifth term is:
(54)(2x)1(−4)4=2560x\binom{5}{4}(2x)^1(-4)^4 = 2560x(45)(2x)1(−4)4=2560x
The final term is:
(55)(2x)0(−4)5=−1024\binom{5}{5}(2x)^0(-4)^5 = -1024(55)(2x)0(−4)5=−1024
Summing these terms, we get:
32x5−320x4+1280x3−2560x2+2560x−102432x^5 - 320x^4 + 1280x^3 - 2560x^2 + 2560x - 102432x5−320x4+1280x3−2560x2+2560x−1024
Therefore, the answer is E.
11112113311464115101051\begin{matrix}& & & & 1 & & & & \\& & & 1 & & 1 & & & \\& & 1 & & 2 & & 1 & & \\& & 1 & & 3 & & 3 & & 1 \\& & 1& & 4 & & 6 & & 4 & & 1 \\& & 1 & & 5 & & 10 & & 10 & & 5 & & 1 \\\end{matrix}11111123451136101410151
The coefficients of the expanded expression will be the entries in the 6th row of Pascal's triangle: 1, 5, 10, 10, 5, 1. We will also have powers of 2x2x2x and powers of −4-4−4 , starting with 5 and decreasing to 0.
Using the row of coefficients and powers, we can build the expanded expression term by term. The first term is:
(2x)5=32x5(2x)^5 = 32x^5(2x)5=32x5
The second term is:
(51)(2x)4(−4)1=−320x4\binom{5}{1}(2x)^4(-4)^1 = -320x^4(15)(2x)4(−4)1=−320x4
The third term is:
(52)(2x)3(−4)2=1280x3\binom{5}{2}(2x)^3(-4)^2 = 1280x^3(25)(2x)3(−4)2=1280x3
The fourth term is:
(53)(2x)2(−4)3=−2560x2\binom{5}{3}(2x)^2(-4)^3 = -2560x^2(35)(2x)2(−4)3=−2560x2
The fifth term is:
(54)(2x)1(−4)4=2560x\binom{5}{4}(2x)^1(-4)^4 = 2560x(45)(2x)1(−4)4=2560x
The final term is:
(55)(2x)0(−4)5=−1024\binom{5}{5}(2x)^0(-4)^5 = -1024(55)(2x)0(−4)5=−1024
Summing these terms, we get:
32x5−320x4+1280x3−2560x2+2560x−102432x^5 - 320x^4 + 1280x^3 - 2560x^2 + 2560x - 102432x5−320x4+1280x3−2560x2+2560x−1024
Therefore, the answer is E.
Learning Objectives
- Implement the principles of the binomial theorem and Pascal's triangle to address challenges in binomial expansion.