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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: 7,935 Thanks: 1129 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: 7,660 Thanks: 2635 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: 20,653 Thanks: 2086 16x² + 8x = (4x+ 1)² - 1 = (4x + 1 - 1)(4x + 1 + 1) = 4x(4x + 2) = 8x(2x + 1) Thanks from SenatorArmstrong and ProofOfALifetime

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