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 July 9th, 2016, 09:51 PM #1 Newbie   Joined: Jul 2016 From: philippines Posts: 7 Thanks: 0 simplifying expressions with negative exponents This is the third time I'm going to ask here and I'm very glad I'm embracing fun times math can give. Still, I have a lot of things to learn like the one above in the title . SO we all know that if we have a negative exponent example a^-2 it will be equivalent to 1/a^2. So what if we are going have equation like this. x(1-2x)^-3/2+(1-2x)^-1/2 how are we going to simplify this thing .... according to the book of Precal in pre-requisite section we will have x/(1-2x)^3/2+1/(1-2x)^1/2...SO according to the book I have red .. to combine the expression you need to get the LCD and by doing so you need to remove the common factor with the smaller exponent..which will give us 1-x/(1-2x)^3/2 ......................so I was like "What the heck did just happen".............. Can you please elaborate...also it is also stated that when factoring we need to subtract exponent.....why subtraction ..why not addition ... so in summary this is my question.. A.How did we arrive at that answer? please elucidate in a detailed way please. B.Why do we need to subtract the exponents, why not add them ? Thank you in advance    July 9th, 2016, 11:14 PM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 I'll address the question of simplifying $\dfrac{x}{(1 - 2x)^{3/2}} + \dfrac{1}{(1 - 2x)^{1/2}}$ first. The first step is to get a common denominator. Note that $\dfrac{1}{(1 - 2x)^{1/2}} = \dfrac{1 - 2x}{(1 - 2x)^{3/2}}$, so our expression becomes $$\dfrac{x}{(1 - 2x)^{3/2}} + \dfrac{1 - 2x}{(1 - 2x)^{3/2}} = \dfrac{1 - x}{(1 - 2x)^{3/2}}$$ As for your question about factorisation,say I wanted to factor the expression $a^2 + a^3$. Obviously, the HCF (highest common factor) is $a^2$, so the factorisation would be $$a^2(\dfrac{a^2}{a^2} + \dfrac{a^3}{a^2}) = a^2(a^{2 - 2} + a^{3 - 2}) = a^2(1 + a)$$ Thanks from topsquark and manus July 10th, 2016, 02:10 AM #3 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 Azzajazz has combined several steps in his working. I recommend beginners writing out every simple single step until they achieve sufficient familiarity to combine steps. So to explain Take $\displaystyle \frac{1}{(1-2x)^{\frac{1}{2}}}$ Multiply by 1 (that's right multiply by one) $\displaystyle \frac{1}{(1-2x)^{\frac{1}{2}}}*1$ But $\displaystyle 1=\frac{1-2x}{1-2x}$ So we have $\displaystyle \frac{1}{(1-2x)^{\frac{1}{2}}}*\frac{1-2x}{1-2x}$ Which equals $\displaystyle \frac{(1-2x)}{(1-2x)^{\frac{3}{2}}}$ Thanks from topsquark and manus July 10th, 2016, 11:07 PM #4 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 Of course, the difficulty is knowing you need to multiply by $\dfrac{1 - 2x}{1 - 2x}$. This is simply a matter of experience. Tags exponents, expressions, negative, simplifying 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 GrannySmith Algebra 3 August 25th, 2014 09:04 PM Danicus Algebra 9 March 10th, 2012 07:41 PM DBJKIBA Algebra 1 September 21st, 2009 06:11 AM zain Elementary Math 2 April 14th, 2008 08:08 PM mercutio Algebra 4 April 8th, 2008 06:44 PM

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