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 October 7th, 2018, 02:00 PM #1 Senior Member   Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics Best way to factor this? 16x^2 + 8x Looks like the solution is (4x+ 2)^2 - 1. Do you arrive at this solution by completing the square? When I tried to complete square things started to go south. Any tips? Thanks! October 7th, 2018, 02:34 PM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond $$16x^2+8x=8x(2x+1)$$ Thanks from SenatorArmstrong Last edited by greg1313; October 8th, 2018 at 12:55 AM. October 7th, 2018, 03:02 PM   #3
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Quote:
 Originally Posted by greg1313 $$16x^2+8x=8x(2x+1)$$

My apologies. I meant to type (4x+1)^2 - 1. Wolfram confirms that an an alternate solution. Is there a technique that can be used to arrive at this?

Last edited by greg1313; October 8th, 2018 at 12:56 AM. October 7th, 2018, 03:19 PM #4 Senior Member   Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics Seems like I need to "force" the expression into a perfect square trinomial. Which is done by adding 1 to both sides. Is there a "trick" to knowing you must add 1 to both sides to turn it into a perfect square trinomial? Or is this just something that comes with practice? Thanks! October 7th, 2018, 03:26 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra \begin{align}16x^2+8x &= (4x)^2 + 2(4x) \\ &= u^2 + 2u &(u=4x) \\ &= \left(u+\tfrac22\right)^2 + c &(\text{constant $c$}) \\ &= \left(u + 1\right)^2 + c \\ &= \left(4x+1\right)^2 + c \\ \cancel{16x^2}+\cancel{8x} &= \cancel{16x^2} + \cancel{8x} + 1 + c \\ 0 &= 1 + c \\ c &= -1\\ \implies 16x^2+8x &= \left(4x+1\right)^2 - 1 \end{align} This isn't a factorisation, it's the vertex form. A factorisation leaves a product of terms as in Greg's post. Thanks from SenatorArmstrong Last edited by v8archie; October 7th, 2018 at 04:12 PM. October 7th, 2018, 04:02 PM   #6
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Quote:
 Originally Posted by v8archie \begin{align}16x^2+8x &= (4x)^2 + 2(4x) \\ &= u^2 + 2u &(u=4x) \\ &= \left(u+\tfrac22\right)^2 + c &(\text{constante $c$}) \\ &= \left(u + 1\right)^2 + c \\ &= \left(4x+1\right)^2 + c \\ \cancel{16x^2}+\cancel{8x} &= \cancel{16x^2} + \cancel{8x} + 1 + c \\ 0 &= 1 + c \\ c &= -1\\ \implies 16x^2+8x &= \left(4x+1\right)^2 - 1 \end{align} This isn't a factorisation, it's the vertex form. A factorisation leaves a product of terms as in Greg's post.
Oh I see! Thanks a lot! Makes sense now! Have a good day!! October 7th, 2018, 05:51 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,967 Thanks: 2216 16x² + 8x = (4x+ 1)² - 1 = (4x + 1 - 1)(4x + 1 + 1) = 4x(4x + 2) = 8x(2x + 1) Thanks from SenatorArmstrong and ProofOfALifetime Tags factor 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 Mukunth Algebra 3 March 9th, 2016 12:08 AM mared Algebra 5 April 23rd, 2014 04:11 AM Eminem_Recovery Algebra 11 June 19th, 2011 08:50 PM haebin Calculus 2 September 14th, 2009 09:25 PM profetas Algebra 3 August 25th, 2009 09:20 AM

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