Consider the Polynomial \(P(x)\). The remainders of \(P(x) \div (x-2)\) is -10 and \(P(x) \div (x+2)\) is \(4\) respectively. Find the remainder of the division \(P(x) \div (x^{2}-4\)
Well, how do i start? I haven't done these in a while
for the first division x-2 = 0 x=2 therefore P(2)=-10
for the second division x+2=0 x=-2 therefore P(-2)=4
for the last division which is ur question actually x^2-4=0 x = +2 or -2 two cases therefore u must take P(+2) & P(-2) both & u know the values of these
the values are remainders actually given by Remainder Theorem.
I think I got it. I assume there's a formula for the remainder?
wait lemme google.
ok:)
So it's \(P(x)=(x-a)q(x)+r(x)\) then \( P(x)-(x-a)q(x) = r(x)\)?
yeah
what is \(q(x)\)?
that is quotient function but i think it is looking more like division lemma
Actual Remainder theorem : - "If f(x) is a function; & (x-a) is its factor then f(a) = 0." If f(x) is a function; & (x-a) is its factor This statement means that f(x) is divisible by (x-a)
how am i able to find the remainder? through assumptions?
i have given that if (x-a) is factor of f(x) then; the remainder will be zero becoz f(a)=0 but if suppose (x-a) is not its factor then the remainder will be f(a) = something but not zero.
which means its \(0\)..?
sorry!
wt do u mean?
i mean wt do u wanna ask actually?
Well, from what you stated. You are saying that the remainder is \(0\)? Which I highly doubt is not.
The remainder is zero when & only when x-a is factor of f(x) or in other words f(a) = 0 otherwise not. for example take ur question for the first division u take x-2 = 0 x=2 but since remainder is -10 then it means that x-2 isn't a factor of P(x) & one more thing P(2) is nt equal to zero since x-2 isn't a factor of P(x) but we know remainder is -10 therefore P(2) =-10
go through my statements clearly u would understand it much better.
alright. thanks
welcome:) but do more questions based on remainder theorem gradually u would understand it:)
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