anonymous
  • anonymous
Factorizing by Quadratics
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[x ^{2}-26x+165\]
DebbieG
  • DebbieG
Look for a factor pair of +165 that sums to -26. Since you want positive constant and a negative coeff on the middle term, you know that you will need both pairs in the factor to be.... ? (both positive? or both negative?) Once you find the right factor pair, a & b, just set up the binomial factors: \(\Large (x\pm a)(x\pm b)\) Just figure out the a, b, and the appropriate signs.
anonymous
  • anonymous
So i will be.. \[\left( x-15 \right) \left( x-11 \right)\] If I am correct?

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DebbieG
  • DebbieG
Yes, very good! :)
anonymous
  • anonymous
Thanks very much :D
anonymous
  • anonymous
Ok this problem is \[x ^{2}-x-20\] Will this one be a bit different?
DebbieG
  • DebbieG
Well, same process. The numbers will be different, of course, lol. Remember, just check factoring with multiplication... ALWAYS.
anonymous
  • anonymous
\[\left( x-19 \right) \left( x-1 \right)\]
DebbieG
  • DebbieG
Welllllllll........ let's check that factoring, by multiplying it back out with FOIL: \(\Large \left( x-19 \right) \left( x-1 \right)=x^2-x-19x+19=x^2-20x+19\) Definitely NOT what you started with.
DebbieG
  • DebbieG
You need FACTOR PAIRS of -20.... remember? What number pairs have a product of 20? (don't worry about the signs, you know they have to be OPPOSITE so that you get a -20 as the last term...) 20=1 * 20 = 2*10 = 4*5 Now which of those, when you use OPPOSITE signs, will SUM to the coefficient of the middle term, which is -1?
anonymous
  • anonymous
Ohh :) \[\left( x-1 \right) \left( x+20 \right)\]
DebbieG
  • DebbieG
did you CHECK BY MULTIPLYING? (I know you didn't... you would have seen: \[\Large \left( x-1 \right) \left( x+20 \right)=x^2 + 19x -20\]
DebbieG
  • DebbieG
READ ABOVE. Which factor pair of 20 (I listed them for you!) will SUM to -1, with the correctly placed opposite signs?
anonymous
  • anonymous
lol :/ *Facepalms* It was so obvious lol. \[\left( x-5 \right) \left( x+4 \right)\]
DebbieG
  • DebbieG
Haha.... yeah, well, we all have those days. YAY, that's it, you got it!
anonymous
  • anonymous
Thanks Very Much :) I became a fan of yours :D
DebbieG
  • DebbieG
Great, thanks, and you're welcome :) Happy to help.
anonymous
  • anonymous
There is one more \[y ^{2} +3y-108\]
anonymous
  • anonymous
I think it will be \[\left( x-9 \right) \left( x+12 \right)\]
DebbieG
  • DebbieG
Good job, that's it!
DebbieG
  • DebbieG
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