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Suppose If $x^2 +px +1$ is a factor of $ax^3 +bx+c$ then relate $a,b,c$ such that $a,b,c \in R$

I can write $$ax^3 +bx+c=(x^2 +px +1)(\lambda x +D)$$ $$\implies ax^3 +bx+c =\lambda x^3 + x^2.p\lambda + x(\lambda+pD)+D $$ and then compare coefficient to find out relation but that will be long and tedious process , I want shorter approach to this problem . Btw I was given following options for this question

A) $a^2+c^2+ab=0$

B) $a^2-c^2+ab=0$

C) $a^2-c^2-ab=0$

D) $ap^2+bp+c=0$

Maybe we can relate something by looking at options?

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3 Answers 3

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Just observe that the product of two roots of the quadratic is $1$. So the third root has to be $\frac{-c}{a}$. Now substitute this root instead of $x$ in the cubic.

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  • $\begingroup$ This is genious , awesome $\endgroup$
    – Tesla
    Mar 25, 2015 at 6:17
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You should have $(x^2+px+1)(\lambda x + D) = \lambda x^3 + (p\lambda + D) x^2 + (pD + \lambda) x + D$. thus $D = c$ and $\lambda = a$, and you need $p \lambda + D = ap + c = 0$ and $pD + \lambda = cp + a = b$. Eliminating $p$ from these two gives you $0 = c (ap + c) - a(cp + a - b) = \ldots$.

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Following the path you have taken, and correcting the algebra, one Gets

$$(\lambda X+D)(X^2+pX+1)=\lambda X^3+(\lambda p+D)X^2+(Dp+\lambda)X+D$$

And this is valid for all $X$. So by identifying the coefficients we get

$$\begin{cases} \lambda=a\\ \lambda p+D=0\\ Dp+\lambda=b\\ D=c \end{cases}$$

which reads

$$\begin{cases} a p+c=0\\ cp+a=b\end{cases}$$

and this leads to

$$-\frac{c^2}{a}+a=b$$

$$a^2-c^2-ab=0$$

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