Question

factor completely: 16x^4-16

Answer 1

Hint: \[\large{a^2-b^2=(a+b)(a-b)}\]

\[\large{(16x^4)-(16)}\]
\[\large{\implies (4x^2)^2-(4)^2}\]

hence put a = \(4x^2\) and b = \(4\)

Answer 2

@Mberdeja501 let me know that what you get after following my instructions.....

Answer 3

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

Answer 4

b = 4 ...
correct it again...

You had put 2 there instead of 4

Answer 5

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

Answer 6

good..
Now, see the second bracket :
\(\large{(4x^2-4) = [{(2x)^2-(2)^2}}]\)

can you again factorize this into (a+b)(a-b) ?

Answer 7

@Mberdeja501 let me know what you get : after following the above expressions

Answer 8

@Mberdeja501 ... r u stuck anywhere ? let me know where r u having problem?

Answer 9

so then i get 2(2x^2+2)(x-1)(x+1)

Answer 10

\[\large{(4x^2+4)(4x^2-4) \implies 4(x^2+1)\times 4(x^2-1)}\]
\[\large{16(x^2+1)(x^2-1) \space \textbf{Factorize x^2-1 now} }\]
\[\large{16(x^2+1)(x+1)(x-1) \space \textbf{(Using a^2-b^2=(a+b)(a-b) identity)}}\]

Answer 11

got it?

Answer 12

@Mberdeja501 ?

Answer 13

uhhhhhmmm, totally lost me

Answer 14

we had this :
\[\large{(4x^2+4) \times (4x^2-4)}\]

correct?