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May 20th, 2015, 06:16 AM  #1 
Member Joined: Mar 2015 From: uk Posts: 33 Thanks: 1  Dividing quadratics
y=(3x^26x10)/(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 rearrange 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 $$\begin{aligned} y &= {3x^26x10 \over 4x^2+8x+5} \\ (4x^2+8x+5)y &= 3x^26x10 \\ (4y3)x^2 + (8y +6)x (5y 10) &= 0 \end{aligned}$$ 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 $x \to \pm\infty$. 
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 prealgebra or Algebra I/II setting. You could just graph it (hint: from 2 to 0).

May 20th, 2015, 06:49 AM  #4 
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  
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  $\displaystyle \frac{3x^26x10}{4x^2+8x+5}$ $\displaystyle =\ \frac34\cdot\frac{x^22x\frac{10}3}{x^2+2x+\frac54}$ $\displaystyle =\ \frac34\left(\frac{[x^2+2x+\frac54]2x2x\frac54\frac{10}3}{x^2+2x+\frac54}\right)$ $\displaystyle =\ \frac34\left(1\frac{4x+\frac{55}{12}}{x^2+2x+\frac54}\right)$ $\displaystyle =\ \frac34\frac{12x+\frac{55}4}{4x^2+8x+5}$ $\displaystyle =\ \frac34\frac{48x+55}{16x^2+32x+20}$ 
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 

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