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Q: 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
A: 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 ]
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User: 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

Weegy: 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 ]
arrabalon|Points 75|

User: 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

Weegy: 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 ]
arrabalon|Points 75|

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Asked 11/8/2012 7:08:18 PM
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