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Factor: y^2 - 4y - 5 Answer: y^2 - 5y - 4y

  1. Jul 7, 2005 #1
    I'm looking over today's chapter and I came across this problem I would like to know if I did this correct.

    The problem is:

    Factor: y^2 - 4y - 5


    y^2 - 5y - 4y
  2. jcsd
  3. Jul 7, 2005 #2
  4. Jul 7, 2005 #3
    To factor a polynomial n is to find two other polynomials a and b such that ab = n. It's the same as number factoring.
  5. Jul 7, 2005 #4
    FOIL(First Outside Inside Last)-how to remember multiplication. reverse it to factor
  6. Jul 7, 2005 #5
    Then the correct answer is:

    x^2 - 4y - 4
  7. Jul 7, 2005 #6
  8. Jul 7, 2005 #7
    where did the x come from?
  9. Jul 7, 2005 #8
    Sorry about that. It is y^2 - 4y -4
  10. Jul 7, 2005 #9
    if you are factoring, your answer will more polynomials than you started with. i do not know how you got the y2-4y-4.
  11. Jul 7, 2005 #10


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    Homework Helper

    We want to write y^2 - 4y - 5 in the form
    (a y+b)(c y+d)=a c y^2+(a d+b c)y+b d
    hence find numbers a,b,c,d such that
    a c=1
    a d+b c=-4
    b d=-5
    First we determine the prime factos of 1 and 5 as the middle term 4 is harder to deal with.
    1 has no prime factors the only way 1 can be written as a product of natural numbers is 1*1 so a=c=1
    5=1*5 so we chose d and b to be either b=-1,d=5 or b=1,d=-5 we guess one and if we are wrong it was the other one.
    lets guess b=-1,d=5
    a d+b c=1*5+(-1)*1=5-1=4 so we guessed wrong
    let b=1,d=-5
    a d-b c=1*(-5)+1*1=-5+1=-4
    right! so
    another way to see this is
    write 4 as 5-1 because 5 is a factor of 5 and 1 is a factor of 1
  12. Jul 7, 2005 #11
    Thank you for the help guys. :smile:
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