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My question is:

Find the value of $k$ such that $$4x^6 - 24x^5 + 20x^4 + 68x^3 -44x^2 - 40x + k$$ is a perfect square.

hey all i have made an edit. Sorry for the inconvenience.

Any help to solve this question would be greatly appreciated.

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The question is incomplete: there is no value of $k$ that makes the polynomial identically equal to $0$, and there are infinitely many values of $k$ that make it equal to $0$ for some $x$. – Brian M. Scott Jun 26 '12 at 8:22
-1 question doesn't make scene – Belgi Jun 26 '12 at 8:24
I presume (based on the question's title) that the goal is to find $k$ such that the polynomial is a perfect square (with integer coefficients). The "$=0$" part is irrelevant. – Blue Jun 26 '12 at 8:24
@meg_1997: Note that all the coefficients we are given are divisible by the square $4$. Divide by $4$. It will make the coefficients pleasantly smaller. – André Nicolas Jun 26 '12 at 9:32
@AndréNicolas: I had thought that too, at first. As it turns out, though, the required $k$ is not divisible by $4$. – Blue Jun 26 '12 at 9:54
up vote 8 down vote accepted

If the goal is to find $k$ so that the polynomial is a perfect square, start by noting that it must be the square of a cubic polynomial:

$$4x^6 - 24x^5 + 20x^4 + 68x^3 -44x^2 - 40x + k=(ax^3+bx^2+cx+d)^2\;.$$

Clearly this immediately require that $a=2$. Now the square of $(2x^3+bx^2+cx+d)^2$ is


so you have the following system of equations:

$$\left\{\begin{align*} &4b=-24\\ &4c+b^2=20\\ &4d+2bc=68\\ &2bd+c^2=-44\\ &2cd=-40\\ &d^2=k \end{align*}\right.$$

Clearly $b=-6$; the second equation then allows you to find $c$, and you can then use the third, fourth, or fifth to find $d$ and then $k$. (To play safe, you should verify that the third, fourth, and fifth equations all yield the same value of $d$.)

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Beat me to it. (I shouldn't've spent time commenting this interpretation of the question!) :) – Blue Jun 26 '12 at 8:37
@DayLateDon: Well, it was the only interpretation that made much sense, given the title! (And the fact that the numbers work out so nicely makes it even more likely, I think.) – Brian M. Scott Jun 26 '12 at 8:40

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