Find a quadratic polynomial which when divided by (x-1), (x-2), (x-3) leaves remainders 11, 22, 37 respectively.

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Hint: Let $f(x)=ax^2+bx+c.$ Then by the given conditions, $$f(x)=(x-1)\phi_1(x)+11$$ $$f(x)=(x-2)\phi_2(x)+22$$ $$f(x)=(x-3)\phi_3(x)+37,$$ for some polynomials $\phi_1(x),\phi_2(x),\phi_3(x)$ of degree $1$.

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Updated on August 18, 2022

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  • pro neon
    pro neon about 1 year

    Find a quadratic polynomial which when divided by $(x-1)$, $(x-2)$, $(x-3)$ leaves remainders $11, 22, 37$ respectively.

    • Naive
      Naive over 6 years
      Show your efforts atleast.
    • lab bhattacharjee
      lab bhattacharjee over 6 years
      Let $f(x)$ $=(x-a)(x-b)(x-1)(x-2)(x-3)$ $\left(\dfrac p{x-a}+\dfrac q{x-b}+\dfrac r{x-1}+\dfrac s{x-2}+\dfrac t{x-3}\right)$
    • Raffaele
      Raffaele over 6 years
      FOUND! It's $P(x)=2 x^2+5 x+4$
    • lioness99a
      lioness99a over 6 years
      We know that the remainder when $f(x)$ is divided by $x-\alpha$ is $f(\alpha)$. If we say that $$f(x)=ax^2+bx+c$$ then we have \begin{align}a\times 1^2+b\times 1+c&=11\\a+b+c&=11\tag{1}\\\\a\times 2^2+b\times 2+c&=22\\4a+2b+c&=22\tag{2}\\\\a\times 3^2+b\times 3+c&=37\\9a+3b+c&=37\tag{3}\end{align} We then have $3$ equations in $3$ unknowns so we can solve these for $a,b,c$: $(2)-(1)$: $$3a+b=11\tag{4}$$ $(3)-(2)$: $$5a+b=15\tag{5}$$ $(5)-(4)$: $$2a=4\\a=2$$ From $(4)$: $$3\times 2+b=11\\b=5$$ From $(1)$: $$2+5+c=11\\c=4$$ And therefore, our quadratic is: $$f(x)=2x^2+5x+4$$
    • lioness99a
      lioness99a over 6 years
      I was too slow to post this as an answer. If it gets reopened, I will convert this to an answer, not a comment
    • pro neon
      pro neon over 6 years
      Thankyou very much sir. You just saved my day. Tomorrow are my exams and I didn't know how to solve this and my teachers don't care about me.
    • lioness99a
      lioness99a over 6 years
      @proneon You're welcome, although I prefer the term 'miss' (as my username may suggest, I am female!) :D
    • pro neon
      pro neon over 6 years
      Sorry about that miss.
  • lab bhattacharjee
    lab bhattacharjee over 6 years