factoring help: 2x^2 - 17x - 100

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
 
got anything against the quadratic formula, jw ? :wink:
 
Yes... the quadratic formula should never be taught (in my opinion) without first understanding how the formula is derived by completing the square. :D
 
factoring help

I cannot figure out if it will factor or not. I am not very good at this math stuff. I just need to learn enough to pass so I can graduate.
 
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. :D
 
kp8179 said:
2x^2-17x-100
Click to expand...
I'm not sure how helpful completing the square or the Quadratic Formula might be for factoring, as opposed to solving...?

You've already received help with this particular exercise, but to learn, in general, how to factor quadratics, try some other web lessons.

Eliz.
 
jwpaine said:
Staple: could you factor this using the grouping method?
Click to expand...
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.

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
 
jwpaine said:
So... A trinomial can't be factored [over the integers] if there are no two factors of [(a)(c)] which add up to = b
Click to expand...
Exactly! :D

Eliz.
 
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