Question

How do you multiply `(x+2)^5`?

Answer 1

`x^5+10x^4+40x^3+80x^2+80x+32`

Explanation:

Use Pascal's triangle (n=5 so 5th row) giving you 1,5,10,10,5,1. The powers of x go from `x^5` down to `x^0` i.e. 1 and the 2nd term contributes `2^0` i.e. 1 up to `2^5`.
So 1st term is 1 (from Pascal) multiplied by `x^5` multiplied by `2^0`
2nd term is 5 multiplied by x^4 then 2^1
and so on.
In an exam, you won't have the triangle so you can use the nCr button to find the multipliers.

Tony B

Answer 2

See a solution process below:

Explanation:

We can use Pascal's triangle to solve this problem.

The triangle values for the exponent 5 are:

`color(red)(1)color(white)(.........)color(red)(5)color(white)(.........)color(red)(10)color(white)(.........)color(red)(10)color(white)(.........)color(red)(5)color(white)(.........)color(red)(1)`

Therefore `(color(blue)(x) + color(green)(2))^5` can be multiplied as:

`color(red)(1)(color(green)(2)^0color(blue)(x)^5) + color(red)(5)(color(green)(2)^1color(blue)(x)^4) + color(red)(10)(color(green)(2)^2color(blue)(x)^3) + color(red)(10)(color(green)(2)^3color(blue)(x)^2) + color(red)(5)(color(green)(2)^4color(blue)(x)^1) + color(red)(1)(color(green)(2)^5color(blue)(x)^0)`

`color(red)(1)(color(green)(1)color(blue)(x)^5) + color(red)(5)(color(green)(2)color(blue)(x)^4) + color(red)(10)(color(green)(4)color(blue)(x)^3) + color(red)(10)(color(green)(8)color(blue)(x)^2) + color(red)(5)(color(green)(16)color(blue)(x)^1) + color(red)(1)(color(green)(32) * color(blue)(1))`

`x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32`

Answer 3

You can use the Binomial Theorem to expand.

Explanation:

`(x+2)^5`

`= ((5),(0)) * x^5 * 2^0 + ((5),(1)) * x^4 * 2^1 + ((5),(2)) * x^3 * 2^2 + ((5),(3)) * x^2 * 2^3 + ((5),(4)) * x^1 * 2^4 + ((5),(5)) * x^0 * 2^5`

`= x^5 + 10x^4 + 40x^3 + 80x^2 +80x + 32`

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By way of explanation using 1 example

`((5),(2))` is another way of writing `color(white)()^5C_2` which is:

`(5!)/((5-2)!2!) = (5xx4xx3!)/((3!)2!) = (5xx4)/(2xx1)xx(3!)/(3!)=10`

The above is called a Combination which in general terms is:
`" "(n!)/((n-r)!r!)`

This is different to Permutations which in general terms is:
`" "(n!)/((n-r)!)`