Factor! 16w^2-56w+49

Question

anonymous · Answer

16w^2-56w+49

akashdeepdeb · Answer

Use (a+b)^2 = a^2 + 2ab + b^2 IDENTITY.

16w^2 - 56w + 49
(4w)^2 - 2(4w)(7) + (7)^2

Understood? :)

NOTE: DO NOT use wolfram alpha.

anonymous · Answer

Is this the answer (4w-7)(4w-7)

akashdeepdeb · Answer

It does but we should work it out ourselves.
It does NOT help in the long run. :)

anonymous · Answer

Im doing it on my notebook thats why. its more easier

akashdeepdeb · Answer

strike out the MORE for correct grammar. :)

akashdeepdeb · Answer

That is the answer. :)

anonymous · Answer

I need help again the problem is a^4-1 and my answer is (a^2-1)(a^2+1). What did i do wrong?

akashdeepdeb · Answer

Factorize  (a^2-1) even further. :)

anonymous · Answer

how

akashdeepdeb · Answer

The same way you factorized (a^4-1).

anonymous · Answer

1 is a prime factor right?

anonymous · Answer

i can't get 1

akashdeepdeb · Answer

See, use the identity (a+b)(a-b) = a2 - b2

Getting the algebraic Identity? :D

anonymous · Answer

(a^2-1)(a^2+2) is this right?

akashdeepdeb · Answer

We have to simplify (a^4-1)

= $$(a^4-1) = (a^2-1)(a^2+1)$$ 
[From Identity (a+b)(a-b) = a2 - b2]

Now good thing is we have $a^2-1$ which we can factorize further using the same identity.

So $a^2-1$ can be written as (a+1)(a-1).

Thus, finally the ultimate factor form of $a^4-1= (a^2+1)(a+1)(a-1)$

Understood? :)

anonymous · Answer

Now i understand

akashdeepdeb · Answer

ok