If the polynomial f (x) = x & # 179; + A & # 178; X & # 178; + AX-1 can be divided by X + 1, then the value of real number a is? The answer is: Because f (x) can be divided by (x + 1), so f (- 1) = - 1 + A & # 178; - A-1 How can f (x) be divided by (x + 1) =

If the polynomial f (x) = x & # 179; + A & # 178; X & # 178; + AX-1 can be divided by X + 1, then the value of real number a is? The answer is: Because f (x) can be divided by (x + 1), so f (- 1) = - 1 + A & # 178; - A-1 How can f (x) be divided by (x + 1) =

F (x) can be divided by (x + 1), which shows that there is (x + 1) in the factor of F (x),
That is, f (x) = (x + 1) (X & # 178; +...) ）Therefore, if f (x) = (x + 1) (X & # 178; +...) ）=When 0, there are (x + 1) = 0 or (X & # 178; +...) )=0,
This shows that the equation f (x) = 0 has a root of - 1, so f (- 1) = 0, that is, f (- 1) = - 1 + A & # 178; - A-1 = 0
Note: your "because f (x) can be divided by (x + 1), so f (- 1) = - 1 + A & # 178; - A-1" should be
Because f (x) is divisible by (x + 1), f (- 1) = - 1 + A & # 178; - A-1 = 0