7
$\begingroup$

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?

$\endgroup$

3 Answers 3

11
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This is genious , awesome $\endgroup$
    – Tesla
    Mar 25, 2015 at 6:17
3
$\begingroup$

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 $0 = c (ap + c) - a(cp + a - b) = \ldots$.

$\endgroup$
0
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.