Christos
  • Christos
Factor 15x^2 - 37x + 20
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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hartnn
  • hartnn
do you know how to generally factor a quadratic ?
Christos
  • Christos
I know some methods but I guess I miss this one
hartnn
  • hartnn
for \(ax^2+bx+c\), you need to find 2 numbers with sum = b and product = ac so, for your question, you need to find 2 numbers with sum =-37 and product = 20*15 = 300

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More answers

Christos
  • Christos
I see now :)
hartnn
  • hartnn
so which 2 numbers u get ? then split -37x accordingly...
Christos
  • Christos
I got -20 and -15 what now?
hartnn
  • hartnn
good! split -37x as -20x-15x so, you have \(15x^2-15x-20x+20\) there's a common factor in 1st 2 terms and a common factor in last 2 terms, can you proceed ?
hartnn
  • hartnn
oh wait, -20 and -15 are incorrect :P
Christos
  • Christos
Yea :S
hartnn
  • hartnn
try another pair ?
Christos
  • Christos
-12 -25 ?
hartnn
  • hartnn
bingo! split -37x as -25x-12x so, you have \(15x^2−25x−12x+20\) there's a common factor in 1st 2 terms and a common factor in last 2 terms, can you proceed ?
Christos
  • Christos
x(x15-25)4(3x + 5)
hartnn
  • hartnn
from first 2 terms factor out 5x
Christos
  • Christos
5x(3x - 5) - 4(3x + 20)
hartnn
  • hartnn
-12x+20 = -4 (3x+.....?)
Christos
  • Christos
its 5 sorry
hartnn
  • hartnn
+5 or -5 ?
Christos
  • Christos
-5
hartnn
  • hartnn
so, now factor 3x-5 from that
Christos
  • Christos
What do you mean? I dont understand :SS
hartnn
  • hartnn
5x (3x-5) -4(3x-5) factor out 3x-5 (3x-5)(5x-4) i meant this, and you're done
anonymous
  • anonymous
Bonjour! Comment allez vous?
hartnn
  • hartnn
i am fine, do u need any help ? with what ?
anonymous
  • anonymous
You are very good at teaching people
hartnn
  • hartnn
oh, thank you very much :) :D
anonymous
  • anonymous
\[15x^2 - 37x + 20 \] \[(3 x-5) (5 x-4)\]

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