Question

without actual division, prove that x^2-4 is a factor of x^4-x^3-6x^2+4x+8​

Answer 1

Answer:

Step-by-step explanation:

since p(x)=x^4 - x^3 - 6x^2 + 4x+8'''''''''''''''''''''''''''''''''''''eq(1)

now x^2-4=0

x^2=4

x=root 4

x=2"""""""""""""""eq(2)

put (2) in (1)

(2)^4 - (2)^3 - 6(2)^2 + 4(2) + 8

16 - 8 - 24 + 8 +8

32-32=0

hence proved that x^2 - 4 is the factor of x^4 - x^3 - 6x^2 + 4x+8

hope that it helps