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May 20th, 2015, 06:16 AM   #1
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Dividing quadratics

y=(3x^2-6x-10)/(4x^2+8x+5)

How do you go about dividing one quadratic into another. I've tried the long division method but end up with terms in x^3, x^4, x^5 ....etc

The question I am looking at is about finding the least upper bound and the greatest lower bound when x is real

The answer is given but no working shown. There is a hint that says re-arrange as a quadratic in x which i assume means turning the above into the form y=ax^2+bx+c

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May 20th, 2015, 06:41 AM   #2
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I suppose that the hint means
$$\begin{aligned}
y &= {3x^2-6x-10 \over 4x^2+8x+5} \\
(4x^2+8x+5)y &= 3x^2-6x-10 \\
(4y-3)x^2 + (8y +6)x (5y -10) &= 0
\end{aligned}$$

The long division should yield $$y = {3 \over 4} - {12x + {55 \over 4} \over 4x^2+8x+5}$$ which is useful for find the limit as $x \to \pm\infty$.
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May 20th, 2015, 06:48 AM   #3
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In general, the two can't be divided -- the division does not result in a polynomial, that is. One way of finding the extrema would be to take the derivative and set it equal to 0, but that's not accessible in a pre-algebra or Algebra I/II setting. You could just graph it (hint: from -2 to 0).
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May 20th, 2015, 06:49 AM   #4
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Quote:
Originally Posted by v8archie View Post
The long division should yield $$y = {3 \over 4} - {12x + {55 \over 4} \over 4x^2+8x+5}$$ which is useful for find the limit as $x \to \pm\infty$.
Yes, but $\pm\infty$ are not extrema in this case.
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May 20th, 2015, 06:51 AM   #5
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$\displaystyle \frac{3x^2-6x-10}{4x^2+8x+5}$

$\displaystyle =\ \frac34\cdot\frac{x^2-2x-\frac{10}3}{x^2+2x+\frac54}$


$\displaystyle =\ \frac34\left(\frac{[x^2+2x+\frac54]-2x-2x-\frac54-\frac{10}3}{x^2+2x+\frac54}\right)$


$\displaystyle =\ \frac34\left(1-\frac{4x+\frac{55}{12}}{x^2+2x+\frac54}\right)$


$\displaystyle =\ \frac34-\frac{12x+\frac{55}4}{4x^2+8x+5}$


$\displaystyle =\ \frac34-\frac{48x+55}{16x^2+32x+20}$
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May 20th, 2015, 01:37 PM   #6
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The least upper bound is 3 and the greater lower bound -3.25. With the values of x that produce these being -5/3 and -5/8

Can someone please tell me how you would get to these answers. I've looked at the working given so far and can't see how to do it

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