factoring help: 2x^2 - 17x - 100
- Thread starter kp8179
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This is hard to intuitively factor... so lets complete the square and then create our binomials. This is long and complicated, but its a straight forward procedure to finding both binomial factors of your trinomial. (once you create them form your zeros)
If you don't know how to complete the square: read this online lesson.
Lets find the zeros
2x^2 - 17x - 100 = 0
2x^2 - 17x = 100
divide out by 2 to get x^2 by itself
x^2 - (17/2)x = 50
take half of the the coefficient b and square it: add this to both sides
x^2 - (17/2)x + (289/16) = 50 + (289/16)
x^2 - (17/2)x + (289/16) = (1089/16)
Convert the left hand side to squared form:
(x - (17/4))^2 = (1089/16)
Solve for your two zeros:
x = (17/4) +/- sqrt( 1089/16 )
x = 25/2 and x = -4
now lets make this into two binomial factors:
(x + 4) and (2x - 25)
If you FOIL (x + 4)(2x - 25) you will get 2x^2 - 17x - 100
If you don't know how to complete the square: read this online lesson.
Lets find the zeros
2x^2 - 17x - 100 = 0
2x^2 - 17x = 100
divide out by 2 to get x^2 by itself
x^2 - (17/2)x = 50
take half of the the coefficient b and square it: add this to both sides
x^2 - (17/2)x + (289/16) = 50 + (289/16)
x^2 - (17/2)x + (289/16) = (1089/16)
Convert the left hand side to squared form:
(x - (17/4))^2 = (1089/16)
Solve for your two zeros:
x = (17/4) +/- sqrt( 1089/16 )
x = 25/2 and x = -4
now lets make this into two binomial factors:
(x + 4) and (2x - 25)
If you FOIL (x + 4)(2x - 25) you will get 2x^2 - 17x - 100
yes, like I showed: you can create two binomial factors, but the process is ugly.
Its not as easy as say, x^2+6x+9
With that.. you know that 3*3 = 9, and 3+3 = 6, so your would factor it into two binomials (x + 3)(x + 3) or, (x+3)^2
Fun stuff.
Its not as easy as say, x^2+6x+9
With that.. you know that 3*3 = 9, and 3+3 = 6, so your would factor it into two binomials (x + 3)(x + 3) or, (x+3)^2
Fun stuff.
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- Feb 4, 2004
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I'm not sure how helpful completing the square or the Quadratic Formula might be for factoring, as opposed to solving...?kp8179 said:
You've already received help with this particular exercise, but to learn, in general, how to factor quadratics, try some other web lessons.
Eliz.
Staple: could you factor this using the grouping method?
http://www.jamesbrennan.org/algebra/pol ... uping_.htm
I tried but cant seem to, so I recreated the binomial factors from the zeros of the polynomial after solving by completing the square.
http://www.jamesbrennan.org/algebra/pol ... uping_.htm
I tried but cant seem to, so I recreated the binomial factors from the zeros of the polynomial after solving by completing the square.
- Joined
- Feb 4, 2004
- Messages
- 16,582
Certainly: Find two factors of (2)(-100) = -200 that add up to -17. (Test pairs, if you're not sure: 1 and 200, 2 and 100, 4 and 50, etc, etc) Split the middle term using these two factors. Factor the resulting four-term polynomial "in pairs", as the site in the link illustrates, and then finish the factorization.jwpaine said:
Where are you stuck? Please be specific. Thank you.
Eliz.
Ok.. I got it.
going down through the only ones I could find are 8 and 25
-25 + 8 = -17
so 8-25
(2x^2+8x)-(25x-100)
remove common factors form each group:
2x(x+4)-25(x+4)
so the grouping method gave me (x+4)(2x-25) (from the left over 2x and -25)
Thanks!
So... A trinomial can't be factored if there are no two factors of [(a)(c)] which add up to = b
going down through the only ones I could find are 8 and 25
-25 + 8 = -17
so 8-25
(2x^2+8x)-(25x-100)
remove common factors form each group:
2x(x+4)-25(x+4)
so the grouping method gave me (x+4)(2x-25) (from the left over 2x and -25)
Thanks!
So... A trinomial can't be factored if there are no two factors of [(a)(c)] which add up to = b