Expand (3x - y)^5 using binomial theorem.
I have a feeling I did this way wrong. Even for such an easy problem...
I get:
243x⁵ - 405(x^4)(y) + 270x³y² - 90x²y³ + 15x(y^4) - y⁵
Let me know if you need working, just a pain to type out!
(3x - y)^5
= ⁵C₀[(3x)^5] - ⁵C₁[(3x)^4y^1] + ⁵C₂[(3x)^3y^2] - ⁵C₃[(3x)^2y^3] + ⁵C₄[(3x)y^4] - ⁵C₅(y^5)
= 1[(3x)^5] - (5/1)[(3x)^4y] + (5 * 4)/(1 * 2)[(3x)^3y^2] - (5 * 4 * 3)/(1 * 2 * 3)[(3x)^2y^3] + (5 * 4 * 3 * 2)/(1 * 2 * 3 * 4)(3xy^4) - 1(y^5)
= 243x^5 - 405x^4y + 270x^3y^2 - 90x^2y^3 + 15xy^4 - y^5
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
I get:
243x⁵ - 405(x^4)(y) + 270x³y² - 90x²y³ + 15x(y^4) - y⁵
Let me know if you need working, just a pain to type out!
(3x - y)^5
= ⁵C₀[(3x)^5] - ⁵C₁[(3x)^4y^1] + ⁵C₂[(3x)^3y^2] - ⁵C₃[(3x)^2y^3] + ⁵C₄[(3x)y^4] - ⁵C₅(y^5)
= 1[(3x)^5] - (5/1)[(3x)^4y] + (5 * 4)/(1 * 2)[(3x)^3y^2] - (5 * 4 * 3)/(1 * 2 * 3)[(3x)^2y^3] + (5 * 4 * 3 * 2)/(1 * 2 * 3 * 4)(3xy^4) - 1(y^5)
= 243x^5 - 405x^4y + 270x^3y^2 - 90x^2y^3 + 15xy^4 - y^5