The quotient when $3 x^5-4 x^4+5 x^3-3 x^2+6 x-8$ is divided by $x^2+x-3$ is

The quotient when $3 x^5-4 x^4+5 x^3-3 x^2+6 x-8$ is divided by $x^2+x-3$ is
  1. $3 x^2-7 x-21$
  2. $3 x^3-7 x^2+21 x-45$
  3. $3 x^4-7 x^3+21 x^2-45+114$
  4. $114 x-143$

Solution

$\begin{aligned} p(x) & =3 x^5-4 x^4+5 x^3-3 x^2+6 x-8 \\ t(x)= & x^2+x-3 \end{aligned}$
By dividing $\mathrm{p}(\mathrm{x})$ by $\mathrm{t}(\mathrm{x})$, the quotient is $3 \mathrm{x}^3-7 \mathrm{x}^2+21 \mathrm{x}-$ 45 and remainder is $114 \mathrm{x}-143$. So, $p(x)=\mathrm{t}(x)\left(3 x^3-7 x^2+21 x-45\right)+114 x-143$ Thus, quotient is $3 x^3-7 x^2+21 x-45$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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