anonymous
  • anonymous
Explain how to factor the following trinomials forms: x2 + bx + c and ax2 + bx + c. Is there more than one way to factor this? Show your answer using both words and mathematical notation.
Mathematics
schrodinger
  • schrodinger
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Hero
  • Hero
Yeah, this is always an interesting topic of discussion. Technically, there's really only one way to factor it. But there are two different approaches to the same thing. Now, however, there are indeed several ways of solving each.
Hero
  • Hero
As you may already know...
Hero
  • Hero
Clearly your instructor must want you to write some long and lengthy explanation for this.

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Hero
  • Hero
:O
myininaya
  • myininaya
ax^2+bx+c I always do: Find two factors of a*c that have product a*c and that have sum b. Say that the we find two factors of a*c and lets call them m and n. Let's assume m and n have product a*c and have sum b. If there are no such m and n then ax^2+bx+c is not factorable over the integers. So if there exist such m and n, then we do the following: Replace bx with mx+nx. We can do this since b=m+n. So now we have ax^2+mx+nx+c. Then we would factor by grouping. (ax^2+mx)+(nx+c) Look to see what the first two terms have in common and then factor that out. Look to see what the second two terms have in common and the factor that out. Then you should have the same binomial times something for each of the two grouping we made. So we would factor this binomial out creating our two factors. Now here is example: 3x^2-2x-5 a=3 b=-2 c=-5 a*c=3(-5) b=-2=3-5 bx=3x-5x 3x^2+3x-5x-5 3x(x+1)-5(x+1) (x+1)(3x-5) here m=a and n=c that is not always the case so how about another example: 12x^2+2x-2 a=12 b=2 c=-2 a*c=12(-2)=-24=(6)(-4) b=2=6-4 bx=2x=6x-4x 12x^2+2x-2 12x^2+6x-4x-2 3x(4x+2)-(4x+2) (4x+2)(3x-1) 2(2x+1)(3x-1) and of course we could had factor out a 2 at the beginning
Hero
  • Hero
Okay, and that's one way to factor....But the other way is still the same way...so there is only ONE way to factor....just two different approaches...and there's still multiple ways of solving...i.e. factoring, quadratic formula, graphing, solving for x, etc...
myininaya
  • myininaya
so at the very beginning of all of this you can say is there a greatest common factor and factor that out at the beginning
myininaya
  • myininaya
greatest common factor of the three terms*
Hero
  • Hero
Also, complete the square but that's just the long form of the quadratic formula.
Hero
  • Hero
Which still isn't factoring

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