How do you expand $(6y^4-1)^2$ ?

Question

2178 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-13T12:19:02

$color(blue)[(6y^4-1)^2=(36*y^8-12y^4+1)]$

Note that:

$color(red)[(a+b)^2=a^2+2ab+b^2]$

$color(red)[(a-b)^2=a^2-2ab+b^2]$

now lets expand $(6y^4-1)^2$

$color(blue)[(6y^4-1)^2=(36*y^8-12y^4+1)]$

Answer 2 · 2018-06-13T12:22:11

$(6y^4-1)^2 = 36y^8 - 12y^4 + 1$

The square of a sum can be expanded as

$(a+b)^2 = a^2+2ab+b^2$

Which means that you have to square both terms, and add twice their product.

The two terms are $6y^4$ and $-1$ . Their squares are $36y^8$ and $1$ .

Their product is $-6y^4$ , so twice their product is $-12y^4$

Now we only need to sum everything together to get

$(6y^4-1)^2 = 36y^8 - 12y^4 + 1$

How do you expand (6y^4-1)^2(6y^4-1)^2?