Question

Q.13. Expand (x^2+ 3y)^5​

Answer 1

Answer:

x^10 + 15(x^8)y + 90(x^6)y²+ 270(x^4)y³ + 405x²y^4 + 243y^5

Step-by-step explanation:

(x²+3y)^5

= x^10 + 5(x^8×3y) + 10(x^6×9y²) + 10(x^4×27y³) + 5(x²×81y^4) + 243 y^5

= x^10 + 15(x^8)y + 90(x^6)y²+ 270(x^4)y³ + 405x²y^4 + 243y^5

Answer 2

Given,

Equation = $(x^2+ 3y)^5$

To Find,

The expanded form of the given equation =?

Solution,

By using the binomial theorem,

$(x^2+3y)^5 = 5C0*(x^2)^{5-0}*(3y)^0 + 5C1*(x^2)^{5-1}*(+3y)^{^1} + 5C2*(x^2)^{5-2}*(3y)^{^2} + 5C3*(x^2)^{5-3}*(3y)^{^3}+ 5C4*(x^2)^{5-4}*(3y)^{^4} + 5C5*(x^2)^{5-5}*(3y)^{5}$

$(x^2+3y)^5= x^{10} + 5(x^{8}*3y) + 10(x^6*9y^2) + 10(x^4 *27y^3) + 5(x^2*81y^4) + 243 y^5\\(x^2+3y)^5= x^{10} + 15(x^8)y + 90(x^6)y^2+ 270(x^4)y^3 + 405x^2y^4 + 243y^5$

Hence, the expanded form of  $(x^2+ 3y)^5$ is $x^{10} + 15x^8y + 90x^6y^2+ 270x^4y^3 + 405x^2y^4 + 243y^5$