When (x^{3}-2 x^{2}+p x-q) is divided by (x^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $f(x) = x^3 - 2x^2 + px - q$ and $g(x) = x^2 - 2x - 3$.First,factorize the divisor: $x^2 - 2x - 3 = (x-3)(x+1)$.Perform polynomial long divisi

Vedclass · Accepted Answer

$p=-2, q=6$

Vedclass · Answer

(C) Let $f(x) = x^3 - 2x^2 + px - q$ and $g(x) = x^2 - 2x - 3$.First,factorize the divisor: $x^2 - 2x - 3 = (x-3)(x+1)$.Perform polynomial long division of $(x^3 - 2x^2 + px - q)$ by $(x^2 - 2x - 3)$:$x^3 - 2x^2 + px - q = x(x^2 - 2x - 3) + (px + 3x - q) = x(x^2 - 2x - 3) + (p+3)x - q$.Thus,the remainder is $(p+3)x - q$.Given that the remainder is $(x-6)$,we equate the coefficients:$(p+3)x - q = 1x - 6$.Comparing the coefficients of $x$ and the constant terms:$p + 3 = 1 \Rightarrow p = -2$.$-q = -6 \Rightarrow q = 6$.Therefore,the values are $p = -2$ and $q = 6$.

When $(x^{3}-2 x^{2}+p x-q)$ is divided by $(x^{2}-2 x-3)$,the remainder is $(x-6)$. The values of $p$ and $q$ are:

Similar Questions

If $2x + \frac{1}{2x} = 2$,then what is the value of $\sqrt{2(\frac{1}{x})^4 + (\frac{1}{x})^5}$?

Resolve into factors: $16(x-y)^{2}-9(x+y)^{2}$

If $x+\frac{1}{x}=2,$ then $x^{3}+\frac{1}{x^{3}}$ is equal to

Resolve into factors: $81 x^{2} y^{2}+108 x y z+36 z^{2}$.

If $x+\frac{1}{y}=1$ and $y+\frac{1}{z}=1,$ find the value of $z+\frac{1}{x}.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module