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Question

(c) Given that ${x^2} + px + 1$is factor of $a{x^3} + bx + c = 0$,

then let $a{x^3} + bx + c \equiv ({x^2} + px + 1)(ax + \lambda )$, where $\lambd

Vedclass · Accepted Answer

${a^2} - {c^2} = ab$

Vedclass · Answer

(c) Given that ${x^2} + px + 1$is factor of $a{x^3} + bx + c = 0$,

then let $a{x^3} + bx + c \equiv ({x^2} + px + 1)(ax + \lambda )$, where $\lambda $ is a constant. Then equating the coefficient of like powers of $x$ on both sides, we get

$0 = ap + \lambda ,\;\;b = p\lambda + a,\;c = \lambda $

$ \Rightarrow p = - \frac{\lambda }{a} = - \frac{c}{a}$

Hence $b = \left( { - \frac{c}{a}} \right)\,c + a$ or $ab = {a^2} - {c^2}$.

If ${x^2} + px + 1$ is a factor of the expression $a{x^3} + bx + c$, then

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