Question

2. Use the binomial theorem to expand the expression. (а — 2b)^5

Answer 1

`(a-2b)^(5)=-32b^(5)+80ab^(4)-80a^(2)b^(3)+40a^(3)b^(2)-10a^(4)b+a^(5)`

Explanation:

The binomial expansion is given by:

`(x+y)^(n)=_(0)^(n)\textrm{C}x^{^(0)}y^(n)+_(1)^(n-1)\textrm{C}x^(1)y^(n-1)+...+_(n)^(n)\textrm{C}x^(n)y^(0)`

In our case we have

`x=a\\y=-2b\\n=5`

Thus using the given terms in the binomial expansion we get

`(a-2b)^(5)=_(0)^(5)\textrm{C}a^(0)(-2b)^(5)+_(1)^(5)\textrm{C}a^{^(1)}(-2b)^(4)+{_(2)^(5)\textrm{C}}a^(2)(-2b)^(3)+_(3)^(5)\textrm{C}a^(3)(-2b)^(2)+_(4)^(5)\textrm{C}a^(4)(-2b)^(1)+_(5)^(5)\textrm{C}a^(5)(-2b)^(0)`

Upon solving we get

`(a-2b)^(5)=-32b^(5)+5* a*16b^(4)+10* a^(2) * (-8b^(3))+10* a^(3)* 4b^(2)+5* a^(4)* (-2b)+a^(5)\\\\(a-2b)^(5)=-32b^(5)+80ab^(4)-80a^(2)b^(3)+40a^(3)b^(2)-10a^(4)b+a^(5)`