My Math Forum Product of Linear Factors

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 January 3rd, 2016, 06:54 PM #1 Newbie   Joined: Sep 2014 From: USA Posts: 6 Thanks: 0 Product of Linear Factors Problem: Rewrite (4x^2+5x)^2 - 5(4x^2+5x) - 6 as a product of linear factors. I have no idea where to start. I tried using the substitution method to get (4x^2 + 5x - 6)(4x^2 + 5x + 1) and using the quadratic formula to get x = 3/4, -2, -1, and -1/4. Last edited by skipjack; January 3rd, 2016 at 07:34 PM.
 January 3rd, 2016, 07:08 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 That's a good approach. If you've got x = 3/4, then 4x - 3 = 0. (4x - 3)(x + 2)(4x + 1)(x + 1) is equivalent to the original expression. Thanks from HowDoIMath Last edited by skipjack; January 3rd, 2016 at 07:29 PM.
January 3rd, 2016, 07:16 PM   #3
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Quote:
 Originally Posted by greg1313 That's a good approach. If you've got x = 3/4, then 4x - 3 = 0. (4x - 3)(x + 2)(4x + 1)(x + 1) is equivalent to the original expression.
Thank you! It was the x = 3/4 that was messing with my answer.
I overlooked 4x - 3 = 0 and used x = 3/4 instead.

Last edited by skipjack; January 3rd, 2016 at 07:30 PM.

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