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A polynomial $f(x) = x^{50} + 3x^{49} + 3x + 12$ when divided by $x - a$, it leaves remainder $3$ & when its quotient is further divided by $x - b$ it leaves remainder $5$, also when $f(x)$ is divided by $x^2 - ( b + a)x + ab$ the remainder is $x + 6$. Find $a$?

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    $\begingroup$ The Answer "a" might come in terms of x as well $\endgroup$
    – Anuj
    Oct 30, 2019 at 13:50
  • $\begingroup$ Where this polynomial belong to? You are considering integer coefficients or complex? $\endgroup$ Oct 31, 2019 at 10:00

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So by Remainder theorem, $f(a)=3$ and $f(x)=g(x)(x-a)(x-b)+(x+6)$ for some polynomial $g(x)$. Using the first result in the second, you immediately get …

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    $\begingroup$ So, by combining the above two answers we get the value of $a$ free from $x$. $\endgroup$ Oct 31, 2019 at 10:12
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$P=x^{50}+3x^{49}+3x+12=x^{49}(x+3)+3(x+3)+3=k(x-a)+3$

So the quotient k is:

$k=\frac{(x^{49}+3)(x+3)}{x-a}$

$\frac{(x^{49}+3)(x+3)}{x-a}=k_1(x-b)+5$

$(x^{49}+3)(x+3)=k_1(x-a)(x-b)+5(x-a)$

$p=(x^{49}+3)(x+3)+3=k_1(x-a)(x-b)+5(x-a)+3$

Therefore:

$5(x-a)+3=x+6$$a=\frac{4x-3}{5}$

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