Question

How do you find the product of `(t+1)(t^2+2t+4)`?

Answer 1

`(t+1)(t^2+2t+4) = t^3+3t^2+6t+4`

Explanation:

You can use the distributive property to find:

`(t+1)(t^2+2t+4) = t(t^2+2t+4)+1(t^2+2t+4)`

`color(white)((t+1)(t^2+2t+4)) = (t^3+2t^2+4t)+(t^2+2t+4)`

`color(white)((t+1)(t^2+2t+4)) = t^3+2t^2+t^2+4t+2t+4`

`color(white)((t+1)(t^2+2t+4)) = t^3+3t^2+6t+4`

Alternatively, just focus on each power of `t` in descending order and add together the different ways of getting it by multiplying a term in the first factor by a term in the second...

`t^3: " "t * t^2 = t^3`

`t^2: " "t(2t)+1(t^2) = 3t^2`

`t: " "t(4)+1(2t) = 6t`

`1: " "1(4) = 4`

Then add:

`(t+1)(t^2+2t+4) = t^3+3t^2+6t+4`

When looking at each power of `t`, we only really need to multiply the coefficients and add the products, so:

`t^3: " "1 * 1 = 1`

`t^2: " "1 * 2 + 1 * 1 = 3`

`t: " "1 * 4 + 1 * 2 = 6`

`1: " "1 * 4 = 4`

With practice you will probably find that you can normally do these calculations in your head as you write down the product:

`(t+1)(t^2+2t+4) = t^3+3t^2+6t+4`