AP EAMCET · Maths · Quadratic Equation
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
- A \(3 x^2-7 x-21\)
- B \(3 x^3-7 x^2+21 x-45\)
- C \(3 x^4-7 x^3+21 x^2-45+114\)
- D \(114 x-143\)
Answer & Solution
Correct Answer
(B) \(3 x^3-7 x^2+21 x-45\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} p(x) & =3 x^5-4 x^4+5 x^3-3 x^2+6 x-8 \\ t(x)= & x^2+x-3 \end{aligned}\) 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,…
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