Question: If (x-2) is a factor of polynomial f(x) but not of g(x), then it must be a factor of? a)f(x) g(x) , b)g(x)-f(x) , c) f(x)-g(x) , d) {f(x)+g(x)}g(x)

Posted by: Suyash D. on 14.03.2021

(x-2) is a factor of f(x), which again is a factor of f(x).g(x). Hence, (x-2) is a factor of f(x).g(x). Thus, answer is option (A) f(x).g(x).
(x-2) is a factor of f(x), which again is a factor of f(x).g(x). Hence, (x-2) is a factor of f(x).g(x). Thus, answer is option (A) f(x).g(x).
A) f(x) g(x) is the only function where x-2 can be taken common from the function as all the other functions contain f(x) either in addition or subtraction to g(x) and thus can't be taken common completely.

Ans. Option a) f(x).g(x)

Reason: f(x).g(x) is the only polynomial given, which already contains the factor (x-2).

Ans: option (a), as (x-2) is a factor of f(x) as well as f(x).g(x) which is divisible by (x-2) although (x-2) is not a factor of g(x).
Ans -a. f (x) g(x)

Answer:  It is given that (x-2) is a factor of  f(x) but not a factor of  g(x). This means that (x-2)  is not a common  factor between  f(x) and g(x). In that case (x-2) can only be a factor of  f(x)*g(x) or f(x) /g(x). Out of the four options given only option (a) satisfies this condition. Hence the correct  answer is option (a).

If x-2 is a facor of f(x) but not g(x), it means f(x) must contain (x-2) term in it as

f(x)=(x-2)f1(x) but g(x ) dos not contain any facor of (x-2)

so when f(x) is multiply anyone its alsohas a factor of x-2.

i.e. f(x)g(x) has factor of (x-2) and

so final answer is Option A

if (x -2) is a factor of polynomial f(x) but not of g(x),then it must be a factor of multiplication of f(x) and g(x). i.e f(x)g(x).

• Answer ( A) F(x) * G(x)  for explanation see the attached file.
A) f(x)g(x) F(x) can be written as (x-2) h(x) So f(x)g(x) =(x-2)h(x)g(x) So f(x)g(x)/(x-2) = (x-2)h(x)g(x)/(x-2) = h(x)g(x) Divisible by (x-2)

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