anonymous
  • anonymous
(x-5)(x+3)=33 ; find the real roots
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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cwrw238
  • cwrw238
(x-5)(x+3)=33 x^2 - 2x - 15 = 33 x^2 - 2x - 48 = 0 can u continue from here?
anonymous
  • anonymous
Thanks! Just realized my mistake. added 33 instead of subtracting
cwrw238
  • cwrw238
yw

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anonymous
  • anonymous
Ah another question. Havent done this stuff in 4 years.
anonymous
  • anonymous
(x+2)^2(x-1)+(x+2)(x-1)^2
whpalmer4
  • whpalmer4
\[(x+2)^2(x-1) + (x+2)(x-1)^2\]multiply it out, step by step: \[(x+2)^2 = (x+2)(x+2) = x^2+2x+2x+4 = x^2+4x+4\] \[(x^2+2x+4)(x-1) = x^3-x^2+2x^2-2x+4x-4 = x^3+x^2+2x-4\] \[x^3+x^2+2x-4 + (x+2)(x-1)^2\]you do the rest, then simplify
anonymous
  • anonymous
I've gotten down to 2x^3+3x^2-11x-2 Any help factoring that please?
whpalmer4
  • whpalmer4
@cwrw238 pretty sure that those 2s outside parentheses are supposed to be ^2
cwrw238
  • cwrw238
yea - u r right
cwrw238
  • cwrw238
i'm embarrassed!!
whpalmer4
  • whpalmer4
life would be easier if openstudiers couldn't copy and paste the questions, losing all of those valuable positional clues :-)
cwrw238
  • cwrw238
yea
anonymous
  • anonymous
Guys , you didnt need to expand!
whpalmer4
  • whpalmer4
we didn't need to do anything, really :-)
anonymous
  • anonymous
just do something similar like a^2v +ab^2 = ab(a+b) then solve!
anonymous
  • anonymous
gah im hitting myself >:(

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