The question is:

Find a cubic polynomial $p(x)$ whose zeroes are the same as those collectively of polynomials $g(x) = 2x^2 - 9x + 4$ and $f(x) = 2x^2 + 3x - 2$. Given that $p(0)$ = 8.

I tried solving the question but I got a little confused at the "collectively" part and also on how to use the value of $p(0)$.

Using the quadratic formula, I calculated the roots of $g(x)$ as 4 and 1/2. Similarly, the roots for $f(x)$ came out to be 1/2 and -2.

And then I added them up to get two roots of the cubic polynomial $p(x)$ as 9/2 and -3/2. ( because they said it's roots are same as those collectively of the two quadratic polynomials. )

I don't exactly know how to proceed after this.

Please explain. Thanks.

  • $\begingroup$ When you collect the root, it means that the roots of $p(x)$ are the same as the roots of $f(x)$ and $g(x)$. Since they are $1/2$,$4$,$1/2$ and $-2$ and that $p(x)$ is cubic, then its roots are $1/2$,$4$ and $-2$. $\endgroup$ – Claude Leibovici Jun 30 '14 at 10:01
  • 1
    $\begingroup$ Thanks a lot. It worked for me. :) $\endgroup$ – EuclidAteMyBreakfast Jun 30 '14 at 10:08
  • $\begingroup$ You are welcome. I am always glad when I can help. I suppose that now you understand what we mean when we $collect$. Cheers :) $\endgroup$ – Claude Leibovici Jun 30 '14 at 10:31
  • $\begingroup$ Yes, I know now. ;) $\endgroup$ – EuclidAteMyBreakfast Jun 30 '14 at 10:36


We have

$$p(x)=\lambda(x-4)(x+2)\left(x-\frac12\right)$$ where $$p(0)=\lambda\times(-4)\times 2\times\left(-\frac12\right)$$

  • $\begingroup$ I didn't get why you put the lambda in there. I managed to get the answer by writing $p(x) = (x - 4) (x + 2) (x - 1/2)$, which led me to the polynomial $2x^3 - 5x^2 - 14x + 8$. Since after collecting the roots, the total number of roots was 3, I don't understand the relevance of giving the value of $p(0)$. Can you please elaborate a little? $\endgroup$ – EuclidAteMyBreakfast Jun 30 '14 at 10:11
  • $\begingroup$ If $p(x)$ is a polynomial that has the roots $4,-2$ and $\frac12$ so what are the roots of $\lambda p(x)$? $\endgroup$ – user63181 Jun 30 '14 at 10:13
  • $\begingroup$ But we have to calculate the roots of $p(x)$, not lamba $p(x)$, right? Or am I missing something? $\endgroup$ – EuclidAteMyBreakfast Jun 30 '14 at 10:15
  • $\begingroup$ For all $\lambda\ne0$, $\lambda p(x)$ has the same roots of $p(x)$ but there's only one polynomial that further has the condition $p(0)=8$ so we should find the adequate $\lambda$ that gives this condition. $\endgroup$ – user63181 Jun 30 '14 at 10:20

We could try $$p(x)=f(x)g(x).$$ At least every root of $f$ or $g$ would then also be a root of $p$. However, this has degree four, not three. Fortunately, $f$ and $g$ have a root in common (found by explicitly determinig the roots or by computing their gcd with Euclid's algorithm), namely $x=\frac12$. So divide by the corresponding linear factor: $$p(x)=\frac{f(x)g(x)}{x-\frac12} $$ Now the only porblem remaining is a scaling factor. Multiply with a suitable constant to ensure $p(0)=8$.

  • $\begingroup$ I think the calculation required to compute this would be long and error-prone. So, in my opinion, it would be better if we just calculate $p(x)$ as $(x - 4) (x + 2) (x - 1/2)$, since the total number of roots is 3 because "1/2" is common, as you rightly said. Also, I don't get what you meant by a "scaling factor". Can you please explain? $\endgroup$ – EuclidAteMyBreakfast Jun 30 '14 at 10:14

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