Question

How do you factor `x^4-44x^2-245`?

Answer 1

`x^4-44x^2-245`

`=(x-7)(x+7)(x^2+5)`

`=(x-7)(x+7)(x-sqrt(5)i)(x+sqrt(5)i)`

Explanation:

We use the difference of squares identity which may be written:

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

So:

`x^4-44x^2-245`

`=(x^2-22)^2-22^2-245`

`=(x^2-22)^2-(484+245)`

`=(x^2-22)^2-729`

`=(x^2-22)^2-27^2`

`=((x^2-22)-27)((x^2-22)+27)`

`=(x^2-49)(x^2+5)`

`=(x^2-7^2)(x^2+5)`

`=(x-7)(x+7)(x^2+5)`

If we allow Complex coefficients:

`=(x-7)(x+7)(x^2-(sqrt(5)i)^2)`

`=(x-7)(x+7)(x-sqrt(5)i)(x+sqrt(5)i)`

Answer 2

`(x^2 + 5)(x - 7)(x + 7)`

Explanation:

Another way to factor the expression -->
Call `x^2 = X` and factor the trinomial:
`y = X^2 - 44X - 245.`
Find 2 numbers knowing sum (-44) and product (-245). They have opposite signs because ac < 0,
Compose factor pairs of (-245) --> (-5, 49)(5, -49). This last sum is
(-44 = b). Then, the numbers are 5 and -49
y = (X + 5)(X - 49) . Replace X by `x^2` -->
`y = (x^2 + 5)(x^2 - 49)`
`y = (x^2 + 5)(x - 7)(x + 7)`