anonymous
  • anonymous
how do you simplify 2a^2+5a-3??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
may be it means to put it into its counterpart... y = M(x-h)^2 +k ?
anonymous
  • anonymous
well the whole equation is 16a^4-1/2a^2+5a-3 * 2a^2+9a+9/4a^2+1
anonymous
  • anonymous
You have to factor it: \[2a^2+5a-3\] You get \[(2a^2)(-a-3)\] Then: \[2a(a+3) -1(a+3)\] Finally: \[(2a-1)(a+3)\]

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amistre64
  • amistre64
\[\frac{16x^4 -1}{2a^2 +5a -3} * \frac{2a^2 +9a +9}{4a^2 +1}\]
anonymous
  • anonymous
you people are a bit confusing- i mean that you're making it more complicated
anonymous
  • anonymous
Yeah, amistre, that's totally irrelevant. It's a quadratic equation; all you have to do is factor it.
anonymous
  • anonymous
so far I have (4a^2+1)(2a+1)(2a-1)/(2a+?) * (2a+9)(a+1)/(4a^2+a)
amistre64
  • amistre64
lol....well it helps to know what the actual question is :)
amistre64
  • amistre64
how do YOU simplify 2x^2 +5a -3 ?? lol
anonymous
  • anonymous
yeah lol that was my question
anonymous
  • anonymous
yes, but not is you're making it more complicated you;re just adding mroe steps
anonymous
  • anonymous
She/he asked for \[2a2+5a−3\] to be simplified. That means factoring it into binomials.
anonymous
  • anonymous
she
amistre64
  • amistre64
i didnt add any steps... I simply was wondering what it was talking about to begin with. right?
anonymous
  • anonymous
\[2a^2+5a−3\] should be the equation; my bad.
anonymous
  • anonymous
I just need that to be factored so I can simplify the rest of the problem
anonymous
  • anonymous
Please read the steps that I posted previously, and tell me if it helps at all.
anonymous
  • anonymous
not really. It only makes it more confusing.
amistre64
  • amistre64
in the context of the rest of the "question", then yes; factoring it is appropriate.
amistre64
  • amistre64
i dont see it factoring to anything useful; so factor the difference of cubes and leave that part alone ;)
anonymous
  • anonymous
Ok thanks :D
amistre64
  • amistre64
you get square roots involved if you factor that thing there, and the other one is clean of them....so they dont cancel anyways... make sense?
anonymous
  • anonymous
A little more than it did. This is what I got: (2a-1)(2a+1)(2a+9)(a+1)/2a^2+5a-3
amistre64
  • amistre64
without doing the math myself, that looks about right :)
anonymous
  • anonymous
Awesome lol thank you
amistre64
  • amistre64
x = -5/4 +- 7/4 x = 12/4 = 3 x = -2/4 = -1/2 (2x-1)(x+3) .... or this :)
amistre64
  • amistre64
(x+6/2)(x-1/2) (x+3)(2x-1) is right... now that I have a moment to see it correctly ;)
anonymous
  • anonymous
That's what I thought it was. Except I had the signs switched.
amistre64
  • amistre64
(2a-1)(2a+1)(2a+9)(a+1) --------------------- (2a-1)(a+3)
amistre64
  • amistre64
2a^2 +9a +9 (a+6/2)(a+3/2) (a+3)(2a+3) is what I get on the top over there
anonymous
  • anonymous
so the answer would be: \[(2a+1)(2a+9)(a+1)\div (a+3)\]
amistre64
  • amistre64
(a+3)(2a+3) (4x^2+1) (4x^2-1) ----------- ---------------- (2a-1)(a+3) (4x^2+1) (2a+3) (4a^2-1) --------------- (2a-1) (2a+3) (2a-1)(2a+1) --------------- (2a-1) (2a+3)(2a+1)
anonymous
  • anonymous
how did you get a+3 on the top?
amistre64
  • amistre64
4a^2 +8a +3 is what I get if I did it right ;) (a+3)(2a+3) = 2a^2 +9a +9
amistre64
  • amistre64
(a+6/2)(a+3/2) (a+3) (a+3/2) (a+3)(2a+3)
anonymous
  • anonymous
but when you factor it back to 2a^2+9a+9...the 3's don't add up to 9
amistre64
  • amistre64
2aa + 3a +2(3)a + 9 2a^2 +3a +6a +9 2a^2 +9a +9
anonymous
  • anonymous
Ooooh ok I think I got
amistre64
  • amistre64
am I right :)
anonymous
  • anonymous
Yup!! Thanks! Now I have 4 more to go....jeez
amistre64
  • amistre64
lol good luck ;)
anonymous
  • anonymous
thanks

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