Question

How do you find the product of `(2t^2+t+3)(4t^2+2t-2)`?

Answer 1

Please see the explanation.

Explanation:

The method for performing the multiplication will become obvious when we write it in a form where each term of the first factor multiplies the second factor:

`(2t^2+t+3)(4t^2+2t-2) =`

`2t^2(4t^2+2t-2) +`

`t(4t^2+2t-2) +`

`3(4t^2+2t-2)`

Distribute the first term of the first factor into the second factor:

`(2t^2+t+3)(4t^2+2t-2) =`

`8t^4+4t^3-4t^2 +`

`t(4t^2+2t-2) +`

`3(4t^2+2t-2)`

Distribute the second term of the first factor into the second factor:

`(2t^2+t+3)(4t^2+2t-2) =`

`8t^4+4t^3-4t^2 +`

`color(white)(".........")4t^3+2t^2-2t +`

`3(4t^2+2t-2)`

Distribute the third term of the first factor into the second factor:

`(2t^2+t+3)(4t^2+2t-2) =`

`8t^4+4t^3-4t^2 +`

`color(white)(".........")4t^3+2t^2-2t +`

`color(white)(".................")12t^2+6t-6`

I have aligned the columns so that they can be added:

`(2t^2+t+3)(4t^2+2t-2) =`

`8t^4+4t^3-4t^2 +`

`color(white)(".........")4t^3+2t^2-2t +`

`ul(color(white)(".................")12t^2+6t-6)`
`8t^4+8t^3+10t^2+4t-6`

Because multiplication is commutative, we could have multiplied the first factor by each term of the second factor and obtained the same result.