If x^2 + px + 1 is a factor of ax^3 + bx + c, | vedclass

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

Question

(B) Since $x^2 + px + 1$ is a factor of $ax^3 + bx + c$,we perform polynomial division:$ax^3 + bx + c = (x^2 + px + 1)(ax - ap) + (b - a + ap^2)x + (a

Vedclass · Accepted Answer

$a^2 - c^2 = ab$

Vedclass · Answer

(B) Since $x^2 + px + 1$ is a factor of $ax^3 + bx + c$,we perform polynomial division:$ax^3 + bx + c = (x^2 + px + 1)(ax - ap) + (b - a + ap^2)x + (ap + c)$For $x^2 + px + 1$ to be a factor,the remainder must be zero:$1) \ ap + c = 0 \Rightarrow p = -\frac{c}{a} \dots (i)$$2) \ b - a + ap^2 = 0$Substituting $(i)$ into the second equation:$b - a + a(-\frac{c}{a})^2 = 0$$b - a + a(\frac{c^2}{a^2}) = 0$$b - a + \frac{c^2}{a} = 0$Multiplying by $a$:$ab - a^2 + c^2 = 0$$a^2 - c^2 = ab$

If $x^2 + px + 1$ is a factor of $ax^3 + bx + c$,then

Similar Questions

If $\sinh(\log x) = -2$,then $x =$

The roots of $4x^2 + 6px + 1 = 0$ are equal,then the value of $p$ is

Let $r(x)$ be the remainder when the polynomial $x^{135}+x^{125}-x^{115}+x^5+1$ is divided by $x^3-x$. Then,

The equation $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ for $x \in R$ has:

For which interval of values of $a$ will the equation $3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0$ have roots of opposite signs?

Vedclass Test Series

Exam Paper Generator

Online Exam Module