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 Algebra Pre-Algebra and Basic Algebra Math Forum

 May 20th, 2015, 06:16 AM #1 Member   Joined: Mar 2015 From: uk Posts: 33 Thanks: 1 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 Thanks May 20th, 2015, 06:41 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 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 $x \to \pm\infty$. Thanks from wirewolf May 20th, 2015, 06:48 AM #3 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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). Thanks from wirewolf May 20th, 2015, 06:49 AM   #4
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Quote:
 Originally Posted by v8archie 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. May 20th, 2015, 06:51 AM #5 Senior Member   Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra $\displaystyle \frac{3x^2-6x-10}{4x^2+8x+5}$ $\displaystyle =\ \frac34\cdot\frac{x^2-2x-\frac{10}3}{x^2+2x+\frac54}$ $\displaystyle =\ \frac34\left(\frac{[x^2+2x+\frac54]-2x-2x-\frac54-\frac{10}3}{x^2+2x+\frac54}\right)$ $\displaystyle =\ \frac34\left(1-\frac{4x+\frac{55}{12}}{x^2+2x+\frac54}\right)$ $\displaystyle =\ \frac34-\frac{12x+\frac{55}4}{4x^2+8x+5}$ $\displaystyle =\ \frac34-\frac{48x+55}{16x^2+32x+20}$ Thanks from topsquark, Country Boy and wirewolf May 20th, 2015, 01:37 PM #6 Member   Joined: Mar 2015 From: uk Posts: 33 Thanks: 1 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 Thanks Tags dividing, quadratics Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post LucilleLelant Algebra 2 May 5th, 2015 08:45 AM Paul9999 Algebra 2 April 15th, 2015 01:31 PM jessjans11 Algebra 3 August 18th, 2014 06:19 AM grangeeducation Algebra 1 April 10th, 2014 09:12 AM sallyyy Algebra 3 June 4th, 2011 01:58 AM

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