Question

# Given f(x) is a cubic polynomial in x. If f(x) is divided by (x+3),(x+4),(x+5) and (x+6) then it leaves the remainders 0,0,4 and 6 respectively. Find the remainder when f(x) is divided by x+7.

A
0
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B
1
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C
2
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D
3
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Solution

## The correct option is A 0It is given that f(x) leaves the remainder 0 if divided by (x+3) and (x+4), which implies that (x+3) and (x+4) are factors of f(x).Let the other factor be ax+p, then f(x) is given by:f(x)=(x+3)(x+4)(ax+p)Now, it is also given that f(x) leaves the remainders 4 and 6 if divided by (x+5) and (x+6), which means that f(−5)=4 and f(−6)=6.If f(−5)=4, then we have:f(x)=(x+3)(x+4)(ax+p)⇒f(−5)=(−5+3)(−5+4)(a(−5)+p)⇒4=(−2)(−1)(−5a+p)⇒4=2(−5a+p)⇒−5a+p=2........(1)And if f(−6)=6, then we have:f(x)=(x+3)(x+4)(ax+p)⇒f(−6)=(−6+3)(−6+4)(a(−6)+p)⇒6=(−3)(−2)(−6a+p)⇒6=6(−6a+p)⇒−6a+p=1........(2)Subtract eqn 2 from eqn 1 as follows:[−5a−(−6a)]+(p−p)=2−1⇒(−5a+6a)+0=1⇒a=1Substitute the value of a in eqn 1:(−5×1)+p=2⇒−5+p=2⇒p=2+5=7Therefore, f(x)=(x+3)(x+4)[(1×x+7)]⇒f(x)=(x+3)(x+4)(x+7)Thus, f(−7)=0Hence, f(x) leaves the remainder 0 when divided by x+7.

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