Friday, 5 June 2015

Factor Theorem
"If P(k)= 0 then x-k is a factor of the polynomial P(x)" and the other way around too: "If x-k is the factor of the polynomial P(x), then P(k)=0".
Proof:
If polynomial P(x) has factor (x-k), then (x-k) must divide P(x), in other words
P(x)= (x-k).H(x)+0, with H(x) is quotient of P(x). 
To prove the theorem above factors, we need to prove first premise to second premise vice versa 
Premise 1:  polynomial P(x) has factor (x-k)
Premis 2: P (k)= 0
1.       If polynomial P(x) has factor (x-k)
for x=k, then
P(k)= (k-k).H(k)+0
P(k)= 0.H(k)+0
P(k)= 0
Proof that if polynomial P(x) has factor (x-k) then P (k)=0
2.       If P(k)= 0
P(x)= A. H(x)              (Suppose A factor of P(x)
P(k)=A. H(k)=0
A.H(k)=0
In order for the above equation right worth 0, the value of A must 0 too.
The the factor of P(x) must (x-k)
According (1) and (2), proof that "If P(k)= 0 then x-k is a factor of the polynomial P(x)" and the other way around too: "If x-k is the factor of the polynomial P(x), then P(k)=0".

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