PROOF 24
Prove.
If f(a) = 0, then (x−a) is a factor
On polynomials.
A way to read this proof
First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.
Definition D1
That (x − a) is a factor of f means there exists a polynomial q with f(x) = (x − a)q(x).
Proof.
A factor is that which divides with no remainder.convention
∎
Lemma L1
The remainder when f is divided by (x − a) is f(a).
Proof.
(1)There exist a polynomial q and a constant r with f(x) = (x − a)q(x) + r.A1
(2)Substitute x = a on both sides.(1), A5
(3)f(a) = (x − a)(a) · q(a) + r(a)(2), A2, A3
(4)(x − a)(a) = a − a = 0A2, A4, A6
(5)r is a constant, so r(a) = rA4
(6)f(a) = 0 × q(a) + r = 0 + r(3), (4), (5), A7
(7)f(a) = r(6), A8
∎
Theorem T
f(a) = 0 holds if and only if (x − a) is a factor of f.
Proof.
(1)Suppose (x − a) is a factor of f.assumption
(2)There is a q with f(x) = (x − a)q(x).(1), D1
(3)Substituting gives f(a) = (x − a)(a) · q(a)(2), A5, A3
(4)(x − a)(a) = 0, so f(a) = 0 × q(a) = 0(3), A2, A4, A6, A7
(5)Conversely, suppose f(a) = 0.assumption
(6)f(x) = (x − a)q(x) + r and r = f(a).A1, L1
(7)r = 0(5), (6)
(8)f(x) = (x − a)q(x) + 0 = (x − a)q(x)(6), (7), A8
(9)Therefore (x − a) is a factor of f.(8), D1
∎
Remark
Can A1 be removed from this proof?
Look at the two directions of the theorem separately. One uses only the definition and substitution; the other does not.