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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.

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  • $\begingroup$ 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 $-2$. $\endgroup$ Jun 30, 2014 at 10:01
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    $\begingroup$ Thanks a lot. It worked for me. :) $\endgroup$ Jun 30, 2014 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 :) $\endgroup$ Jun 30, 2014 at 10:31
  • $\begingroup$ Yes, I know now. ;) $\endgroup$ Jun 30, 2014 at 10:36

3 Answers 3

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Hint

We have

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

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  • $\begingroup$ 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? $\endgroup$ Jun 30, 2014 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, 2014 at 10:13
  • $\begingroup$ But we have to calculate the roots of $p(x)$, not lamba $p(x)$, right? Or am I missing something? $\endgroup$ Jun 30, 2014 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, 2014 at 10:20
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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 $p(0)=8$.

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  • $\begingroup$ 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? $\endgroup$ Jun 30, 2014 at 10:14
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It is also possible to work out the coefficients of the cubic polynomial without using the quadratic formula on the two given polynomials (although we'll find that we pick up all three roots along the way). Since neither of the quadratic polynomials is a "binomial-square", we infer that the three roots of the cubic polynomial are distinct; we are also given the constant term for the cubic, so we have

$$ ax^3 + bx^2 + cx + 8 \ = \ a · (x-r)·(x-s)·(x-t) \ \ , $$ $$ 2x^2 - 9x + 4 \ = \ 2 · (x-r)·(x-s) \ \ , \ \ 2x^2 + 3x - 2 \ = \ 2 · (x-r)·(x-t) \ \ . $$

Viete's relations tell us that $ \ -arst \ = \ 8 \ \ , \ \ 2rs \ = \ 4 \ \ , \ \ 2rt \ = \ -2 \ \ $ (we will have further use for them shortly).

If we subtract one of the quadratic polynomials from the other (convenient here because the leading coefficients of both are the same), we have

$$ 2·(x-r)·(x-s) \ - \ 2·(x-r)·(x-t) \ \ = \ \ 2·(x-r)·[(x-s) \ - \ (x-t)] $$ $$ = \ \ 2·(x-r)·(t-s) $$ $$ \Rightarrow \ \ (2x^2 - 9x + 4) \ - \ (2x^2 + 3x - 2) \ = \ -12x + 6 \ = \ -12·\left(x - \frac12 \right) \ \ . $$ [This is related to the Euclidean algorithm technique mentioned by Hagen von Eitzen .] We now know that the recurring zero is $ \ r = \frac12 \ $ and the difference between the other zeroes is $ \ 2·(t-s) \ = \ -12 $ $ \Rightarrow \ t-s \ = \ -6 \ . $ (We don't end up using this difference directly, but we'll soon see that it is so.)

At this point, we have everything we need to find the three unknown coefficients: $$ 2rs \ = \ 4 \ \ \Rightarrow \ \ rs \ = \ 2 \ \ \Rightarrow \ \ s \ = \ 4 \ \ , \ \ 2rt \ = \ -2 \ \ \Rightarrow \ \ rt \ = \ -1 \ \ \Rightarrow \ \ t \ = \ -2 \ \ , $$ $$ -arst \ = \ 8 \ \ \Rightarrow \ \ a · \left(\frac12 \right) · 4 · (-2) \ = \ -8 \ \ \Rightarrow \ \ a \ = \ 2 \ \ . $$

Viete goes on to tell us that $$ b \ = \ a·(-r-s-t) \ = \ 2 · \left(-\frac12 \ - \ 4 \ - \ [-2] \right) \ = \ -5 \ \ , $$ and $$ c \ = \ a·(rs+rt+st) \ = \ 2 \ · \ \left(2 \ + \ [-1] \ - \ 4·[-2] \right) \ = \ -14 \ \ . $$

Our cubic polynomial is therefore $ \ 2x^2 - 5x^2 - 14x + 8 \ \ . $ [Solving for zeroes by way of an online resource -- or the Rational Zeroes Theorem, synthetic division, and patience -- confirms this result.]

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To elaborate on Hagen von Eitzen's approach, once we have found the common linear factor between the two quadratic polynomials, we can obtain the cubic polynomial from $$ \frac{k \ · \ [2·(x-r)·(x-s)] \ · \ [2·(x-r)·(x-t)]}{(x - r)} \ \ = \ \ \frac{k \ · \ (2x^2 - 9x + 4)· (2x^2 + 3x - 2)}{\left(x - \frac12 \right)} \ \ , $$ where the common factor between the polynomials is found as discussed above and $ \ k \ $ has been referred to as a "scaling factor", which must be brought in to be sure that the given constant term of $ \ 8 \ $ is produced. (The division by that monic linear factor does not affect the leading coefficient.) The cubic polynomial formed by this ratio must have the specified value for $ \ x = 0 \ \ , $ hence,

$$ \frac{k \ · \ (4)· (- 2)}{\left(- \frac12 \right)} \ = \ 8 \ \ \Rightarrow \ \ k \ = \ \frac12 \ , $$

We can now calculate (by hand or with on-line aid)

$$ \frac{\frac12 \ · \ (2x^2 - 9x + 4)· (2x^2 + 3x - 2)}{\left(x - \frac12 \right)} \ \ = \ \ \frac{2x^4 \ - \ 6x^3 \ - \ \frac{23}{2}x^2 \ + \ 15x \ - \ 4}{\left(x - \frac12 \right)} \ \ . $$

We then apply polynomial or synthetic division here (or software):

$$ 1/2 \ \ \ | \ \ 2 \quad -6 \quad -23/2 \quad 15 \quad -4 $$ $$ \quad \quad \quad \quad 1 \quad \ \ -5/2 \quad -7 \quad \quad 4$$ $$ \quad \quad \quad ---------------- $$ $$ \quad \quad 2 \quad -5 \quad \ -14 \quad \quad 8 \quad \quad | \ \ 0 $$

to yield the "reduced" polynomial, $ \ 2x^3 - 5x^2 - 14x + 8 \ \ . $

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