Question

How would I factor this:
16(x^2+1)^2-4(2x)^2

Answer 1

\[\large 16(x^2+1)^2 -4(2x)^2\]Expand \(\large 16\color{blue}{(x^2+1)^2}-4\color{blue}{(2x)^2}\)

What do you get?

Answer 2

16(x^2+1)(x^2+1)-4(2x)(2x)

Answer 3

Yes, that's correct, or you can apply the formula \(\large a^2 +2ab+b^2\) to \(\large x^2+1\) It will result in

\[\large 16\color{blue}{(x^4 +2x^2 +1)}-4\color{blue}{(4x^2)}\]

Answer 4

I have never seen that formula before. can we solve with out it?

Answer 5

16(x^2+1)^2 - 4(2x)^2
= 4^2(x^2+1)^2 - 2^2(2x)^2
= (4(x^2+1))^2 - (2(2x))^2
= (4x^2+4)^2 - (4x)^2

then use the difference square formula :
a^^2 - b^2 = (a+b)(a-b)

continue it ...

Answer 6

* a^2

Answer 7

I guess whichever method you prefer....

Answer 8

Still there?

Answer 9

ya

Answer 10

still trying to figure it out. I think Im suppose to use the difference of squares thing

Answer 11

You can use any method you like, really. It'll all result in the same solution.

Answer 12

Dont understand why 4^2 and 2^2 are in line 2
Dont I take the gcf out

Answer 13

in Raden

Answer 14

To make it to the same power, so the terms can be combined.

Answer 15

\[\large 16(x^2+1)^2 - 4(2x)^2\] because \(\large (x^2+1)^2\) is to the second power, we can change \(\large 16 \iff 4^2\) so these two terms, \(\large (x^2+1)^2\) and \(\large 4^2\) are to the same power. Therefore they can be combined as so. \[\large 4^2(x^2+1)^2 = (4(x^2+1))^2\]

Answer 16

Same thing with \(\large 4(2x)^2\)

$ can be rewritten : \(\large 4 \iff 2^2 \)

So we would have: \(\large 2^2(2x)^2 \iff (2(2x))^2\)

Answer 17

It*

Answer 18

Ok how do I then find the factors. I understand everything p to the last line in rd example

Answer 19

up

Answer 20

So now you want to utilize the formula for difference of squares, \(\large a^2 -b^2\).

You've got: \(\large [4(x^2+1)]^2- [2(2x)]^2\) First expand whats inside the brackets.

\(\large (4x^2+4)^2 -(4x^2)\) this is your difference of squares equation.

\(\large a^2 = (4x^2+4)^2 \ \therefore \ a = 4x^2+4\)

\(\large b^2 = (4x)^2 \ \therefore \ b = 4x\)

Answer 21

the answer in the back said
16(x^2-x+1)(x^2+x+1) I dont see how this works out to be hat

Answer 22

not sure how they came up with that answer

Answer 23

\[\large (a+b)(a-b) = (4x^2+4+4x)(4x^2+4-4x)\]

Answer 24

Now use the foil method or grouping method to solve.

Answer 25

And easy method would be to factor out the GCF, 4.

Answer 26

ok cool

Answer 27

What do you get?

Answer 28

4(x^2+1+x)x^2+1-x)

Answer 29

forgot the bracket in front of the x

Answer 30

You made a small mistake there. We would be factoring out a 4 from each of the two terms, \(\large (4x^2+4+4x)\) and \(\large (4x^2+4-4x)\)

Answer 31

\[\large 4(x^2+x+1) \cdot 4(x^2 -x+1) = (4 \cdot 4)(x^2+x+1)(x^2-x+1) \]

Answer 32

ohh I see

Answer 33

This is a great example I can use a reference. Thanks for your time. How do I save this response ?

Answer 34

are you using chrome web browser?

Answer 35

safari

Answer 36

I got link sent to me in the mail email

Answer 37

Thanks again

Answer 38

Oh ok :)