a friend needs this explained, I can't remember how to do it. It's (y-2)(y+2)(y squared+4)?

Question

algebra, math problem

Answer 1

(y-2)(y+2)(y^2+4)=(y^2-4)(y^2+4)=y^4-16

Answer 2

This is merely a multiplication problem. Learn the FOIL (front, outer, inner, last) technique and all falls into place.

Multiply the first 2:
(y-2)(y+2) = y^2 + 2 y - 2 y -4 = y^2 - 4 (understand this and you have the problem solved)

now
(y^2 -4)(y^2 + 4) = y^4 + 4y^2 -4y^2 - 16 = y^4 - 16

Answer 3

First we lets multiply out (y-2) and (y+2)
y*y+2y-2y-4

We end up with (y^2-4)

ok now lets multiply out (y^2-4)(y^2+4)
Y^4-4y^2+4y^2-16

Lets now simplify y^4-16

now let set up an equation

y^4-16=0

add 16 to both sides

y^4=16

Now lets solve for y
take the 4th root(I do not know the symbol for this) of both sides of the equation

y=plus or minus 2

2*2*2*2=16
-2*-2*-2*-2=16

Anwsar y=2 or -2

Answer 4

(y - 2)(y + 2)(y^2 + 4)

(y^2 - 4)(y^2 + 4)

y^4 - 16

Difference of two squares

a^2 - b^2 = (a + b)(a - b)

Answer 5

(y-2)(y+2)(y^2 +4) = (combine first two)
(y^2 -4 )(y^2 + 4) = multiply directly
y^4 - 16 (it's the difference of two squares)

Answer 6

( y - 2 )( y + 2 )( y^2 + 4 ) = (y^2 - 2y + 2y - 4) ( y^2 +4)
= (Y^2 - 4) (y^2 +4 ) = Y^4 - 4 Y^2 + 4Y^2 - 16
= Y^4 - 16