Question

if f(x) is divided by x-3 , then the remainder is 5 and when it is divided by x-4 , then remainder is '0' . then remainder when it is divided by (x-3)(x-4) is

Answer 1

Answer:

Given, p(x) when divided by x−3 and x−5, leaves remainder 10 and 6, respectively.

From remainder theorem of polynomial, p(3)=10 and P(5)=6.

If the polynomial is divided by (x−3)(x−5), then the remainder must be of the form ax+b (Since, degree of remainder is less than that of the divisor).

⇒p(x)=q(x)(x−3)(x−5)+(ax+b), where q(x) is some polynomial (quotient obtained).

Substituting for x=3 and x=5:

p(3)=q(x)(3−3)(3−5)+a(3)+b=0+3a+b=3a+b⟹10=3a+b−−(i)

p(5)=q(x)(5−3)(5−5)+a(5)+b=0+5a+b=5a+b⟹6=5a+b−−(ii).

Solving for a and b, we get,

(i)−(ii)⟹10−6=(3−5)a⟹4=−2a⟹a=−2−−(a).

(a) in (ii)⟹6=5(−2)+b⟹6=−10+b⟹b=6+10⟹b=16.

∴a=−2 and b=16.

⇒ Remainder =−2x+16.

Answer 2

Answer:

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