Answer:
Step-by-step explanation:
roots are (2+3i) and (2-3i)
reqd. eq. is (x-2-3i)(x-2+3i)=0
or (x-2)²-(3i)²=0
or x²-4x+9-9i²=0
or x²-4x+9+9=0
or x²-4x+18=0
Consider this solution, if it is possible, check the result:
1) if to divide polynomial 'x³-3x²-6x+8' on 'x+2' result is: x²-5x+4, then x³-3x²-6x+8=(x+2)(x²-5x+4);
2) x²-5x+4=(x-1)(x-4), then
3) x³-3x²-6x+8=(x+2)(x-1)(x-4).
4) zeros are: -2;1;4. Other zeros are: 1 and 4.
Answer: 1;4.
Answer:
can u help answer mine
Step-by-step explanation:
Answer:
1325/10000
Explain Step By Step Answer
Answer:
x-2
The choose (1)
Step-by-step explanation:
(x³-3x²+3x-2)÷(x²-x+1)
(x-2)(x²-x+1) ÷ (x²-x+1)
Delete (x²-x+1)
so = (x-2)
