MCQ
If ${x^2} + px + 1$ is a factor of the expression $a{x^3} + bx + c$, then
- A${a^2} + {c^2} = - ab$
- B${a^2} - {c^2} = - ab$
- ✓${a^2} - {c^2} = ab$
- DNone of these
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 $ab = {a^2} - {c^2}$.
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