Can you factor these for me? Please show all steps!?

Follow
  • Follow publicly
  • Follow privately
  • Unfollow
x2-x-6=0 (The 2 is an exponent) x2+5x+4=0 (The 2 is an exponent) x2-4x-5=0 (The 2 is an exponent, they're all going to be that way so you get the point) x2+x-12=0 ...show more
Best Answer
Hi,

For problems starting with x², look at the last sign. If it is positive, find factors of the constant (the last number) that add to the middle number. If the last sign is negative, find
factors of the constant (the last number) that subtract to the middle number.

x2-x-6=0
(x - 3)(x + 2) = 0
x = 3 or -2

x2+5x+4=0
(x + 4)(x + 1) = 0
x = -4 or -1

x2-4x-5=0
(x - 5)(x + 1) = 0
x = 5 or x = -1

x2+x-12=0
(x + 4)(x - 3) = 0
x = -4 or x = 3

x2-3x-10=0
(x - 5)(x + 2) = 0
x = 5 or x = -2

x2+7x+12=0
(x + 4)(x + 3) = 0
x = -4 or x = -3

x2+2x-15=0
(x + 5)(x - 3) = 0
x = -5 or x = 3

x2-9x+14=0
(x - 7)(x - 2) = 0
x = 7 or x = 2

x2-7x-18=0
(x - 9)(x + 2) = 0
x = 9 or x = -2

x2+4x-45=0
(x + 9)(x - 5) = 0
x = -9 or x = 5

x2+9x+18=0
(x + 6)(x + 3) = 0
x = -6 or x = -3

x2-8x+15=0
(x - 5)(x - 3) = 0
x = 5 or x = 3

x2+6x+5=0
(x + 5)(x + 1) = 0
x = -5 or x = -1

I hope that helps!! :-)
  • Rate
  • Comment

Other Answers (8)

Rated Highest
  • Rated Highest
  • Oldest
  • Newest
  • ʎɥʇɐʞ answered 4 years ago
    too many!
    • 2
    • Comment
  • Ash57 answered 4 years ago
    seriously...all of these
    • 2
    • Comment
  • diablonecros answered 4 years ago
    wow if it was only one or 2 its okay but theres soo many. btw just by looking at them, they arent that hard to factor, just too time-consuming
    • 1
    • Comment
  • archimedes answered 4 years ago
    You are being asked questions about them and you are NOT PREPARED !

    These are basic and simple trinomial factoring problems. You should be able to do these. If you really can't do any of these, then you must go to your teacher. Obviously, the concept of using your text book is not in your repertoire !

    You will have to pass tests on this material. I suggest you learn it !

    QED
    • 1
    • Comment
  • peabody answered 4 years ago
    I will teach you how to do this. The answers are in the back of your text book.

    eg. x^2 + 5x +6 = 0

    ( x )( x ) = 0 First what to multiply to get x^2 x * x = x^2

    ( x + 3 )( x +2 ) = 0 What two numbers have a product of 6 but add to 5.
    Answer 3 and 2

    Therefore x+3 =0 or x+2 = 0 So x = -3 or -2

    x^2 - 3x + 10 = 0

    ( x - 5 )( x +2 ) =0 x * x = x^2 , -5 * 2 = -10 , -5 +2 =-3

    Therefore x-5 =0 or x+2 = 0

    so x = 5 or -2
    • Rate
    • Comment
  • Eng Ngee answered 4 years ago
    (x-3)(x+2)

    (x+1)(x+4)

    (x-5)(x+1)

    (x+4)(x-3)

    (x-5)(x+2)

    (x+4)(x+3)

    (x+5)(x-3)

    (x-7)(x-2)

    (x-9)(x+2)

    (x+9)(x-5)

    (x+3)(x+6)

    (x-3)(x-5)

    (x+1)(x+5)
    • Rate
    • Comment
  • Bo Q answered 4 years ago
    x2+5x+4=0=(x+4)(x+1)
    x2-4x-5=0=(x-5)(x+1)
    x2+x-12=0 = (x+4)(x-3)
    x2-3x-10=0=(x-5)(x+2)
    x^2 + x -12 = (x + 4)(x-3)
    x^2 -9x +14 = (x-7)(x-2)
    x2-7x-18=0 = (x-9)(x+2)
    x2+4x-45=0 = (x+9)(x-5)
    x2+9x+18=0 = (x+3)(x+6)
    x2-8x+15=0 = (x-3)(x_5)
    x2+6x+5=0 = (x+5)(x+1)
    • Rate
    • Comment
  • Ashley Jeanne answered 4 years ago
    x^2 - x - 6 = 0 Factor (x +2)(x - 3)
    x^2 + 5x + 4 = 0 Factor (x + 4)(x + 1)
    x^2 + x - 12 = 0 Factor (x + 4)(x - 3)
    x^2 - 3x - 10 = 0 Factor (x - 5)(x + 2)
    x^2 + 7x + 12 = 0 Factor (x+3)(x + 4)
    x^2 + 2x - 15 = 0 Factor (x + 7)(x - 5)
    x^2 - 9x + 14 = 0 Factor (x - 7)(x - 2)
    x^2 - 7x -18 = 0 Factor (x - 9)(x + 2)
    x^2 + 4x - 45 = 0 Factor (x + 9)(x - 5)
    x^2 + 9x + 18 = 0 Factor (x = 6)(x + 3)
    x^2 -8x + 15 = 0 Factor (x - 5)(x - 3)
    x^2 + 6x + 5 = 0 Factor (x = 5)(x + 1)

    OK all done
    • Rate
    • Comment
  • Sign In 

    to add your answer

Who is following this question?

    %
    BEST ANSWERS
    Member since:
    Points: Points: Level
    Total answers:
    Points this week:
    Follow
     
    Unfollow
     
    Block
     
    Unblock