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I am unsure of how to approach this problem. I have thought about using the Rational root theorem, but I am unsure if this answers the question being asked.

Using the theorem, I get $\frac{p}{q} = \pm 1, \pm 13, \pm \frac{1}{5}$, and $\pm \frac{13}{5}$ as possible roots. Then I use synthetic division and Horner's method to get a remainder of $-(n+8)$. For this to be a solution, $-(n+8)=0$, so $n = -8$. Then I could do this for $+1, +13, -13,$ etc.

Is this the correct approach to answering the original question? Original question:

Find all integers $n$ such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients.

Why would I need to have a rational root to answer the problem? Couldn't I have complex solutions where I can express $5x^2 + nx - 13$ (where n is an integer) as a product of two linear factors with integer coefficients?

I greatly appreciate any insight you could provide on this. It's been about 2 years since I've done any mathematics (a brief foray into Neuroscience turned into a longer expedition than intended) and I am longing to return to the beautiful realm of mathematics. Thanks for your time in reading through this jumbled mathematical thought!

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    $\begingroup$ One thing to consider is $b^2-4ac$ or $n^2+260$ needs to be a perfect square. $\endgroup$ – turkeyhundt Apr 24 '15 at 4:27
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    $\begingroup$ The values of n I have calculated are +/- 8 and +/- 64. All four of these n values satisfy the determinant being a perfect square. I am curious of the reason why it must be a perfect square. Why is this? $\endgroup$ – MLiuzzolino Apr 24 '15 at 4:36
  • $\begingroup$ Our roots are of the form $\frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$. Since $a, b$, and $c$ are integers, the above roots will only be rational when $\sqrt{b^2 - 4ac}$ is rational, and this only happens when $b^2 - 4ac$ is a perfect square (integer or rational). $\endgroup$ – pjs36 Apr 24 '15 at 4:53
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More simply, note that $5$ is prime, so the linear factors have to have coefficients of $x$ as $1$ and $5$ (or $-1$ and $-5$).

So, you have $(x-r)(5x-s) = 5 x^2 -nx -13$ for integers $r,s$. FOILING the left hand side gives $5x^2 - (5r+s) x + rs$, so $rs=-13$ and $5r+s=n$. Since $13$ is prime, we must have $r=-1,s=13$ or $r=1,s=-13$ to meet the constraint that $rs=-13$. You can then plug these in and find $n$.

You can do the same thing with $(-x -r) (-5x-s) = 5 x^2 -nx -13$ to get the other case.

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  • $\begingroup$ Batman, thanks! That is definitely a lot simpler and makes sense. Thank you! $\endgroup$ – MLiuzzolino Apr 24 '15 at 4:41
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We want to find $n$ such that $f(x) = 5x^2 +nx -13$ is a product of two linear terms with integer coefficients. Now let $f(x)=(ax+b)(cx+d) = acx^2 +(ad+bc)x +bd $ with $a,b,c,d$ integer. Then $ ac= 5 , bd =-13 , ad+bc =n$. So $(a,c)=(5,1 ),(-5,-1)$ and $(b,d)= (13,-1),(-13,1)$. This implies $n = ad+bc$ can be $\pm 64, \pm 8$

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  • $\begingroup$ There are some type-os ... $(ax+b)(cx+d)=acx^2+(ad+bc)x+bd$ and $bd=-13$ You changed $ad+bc$ to $ac+bd$... And also those n's aren't correct. $\endgroup$ – randomgirl Apr 24 '15 at 4:51
  • $\begingroup$ (a,c)=(5,1),(-1,-5) only since a*c=5 and (b,d)=(13,-1),(-13,1) since bd=-13...and n=ad+bc so you have the following n=5(13)+1(-1) or n=5(-13)+1(1) or n=-1(13)+(-5)(-1) or n=-1(-13)+(-5)(1)... Let me know if I made a mistake @Shouman Das . $\endgroup$ – randomgirl Apr 24 '15 at 5:02
  • $\begingroup$ fixed it. thanks. $\endgroup$ – aNumosh Apr 24 '15 at 5:03
  • $\begingroup$ Except that last n expression but much better. like n=ad+bc . :p $\endgroup$ – randomgirl Apr 24 '15 at 5:03
  • $\begingroup$ So perfect. I gave you +1 already. :) $\endgroup$ – randomgirl Apr 24 '15 at 5:09
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This can be done if and only if $5x^2+nx-13$ is a difference of two squares. $$5x^2+nx-13=\frac15(25x^2+5nx+\left(\frac{n}{2}\right)^2)-(13+\frac{n^2}{4})$$ $$5x^2+nx-13=\frac15(5x+\frac{n}{2})^2-\frac14(n^2+52).$$ Therefore, there should be an integer $m$ such that $$n^2+52=5m^2.$$ I think this will helps you. Continue from here.

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If a quadratic polynomial $ax^2 + bx + c$ with integer coefficients, $a \neq 0$, has factorization $$ax^2 + bx + c = (rx + s)(tx + u) \tag{1}$$ then \begin{align*} a & = rt\\ b & = ru + st\\ c & = su \end{align*} which can be demonstrated by substituting $0$, $1$, and $-1$ in equation 1, then solving the resulting system of equations. Note that $ac = (rt)(su) = rstu = (ru)(st)$, that is, $b$ is the sum of two integers whose product is $ac$.

For the equation $5x^2 + nx - 13$, this means that $n$ is the sum of two integers with product $5 \cdot -13 = -65$. Since \begin{align*} -65 & = 1 \cdot -65 & -65 & = -1 \cdot 65\\ & = 5 \cdot -13 & & = -5 \cdot 13 \end{align*} the possible integer values for $n$ are \begin{align*} 1 + (-65) & = -64 & -1 + 65 & = 64\\ 5 + (-13) & = -8 & -5 + 13 & = 8 \end{align*}

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