How do you factor #x^4-44x^2-245#?
2 Answers
#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)#
Explanation:
Another way to factor the expression -->
Call
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
