Notice: Payments for answers will end 4/10/2017. Click for more info.
You have new items in your feed. Click to view.
Question and answer
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 ]
Get an answer
Original conversation
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|

Question
Asked 11/8/2012 7:08:18 PM
0 Answers/Comments
New answers
Rating

There are no new answers.

Comments

There are no comments.

Add an answer or comment
Log in or sign up first.
26,488,509 questions answered
Popular Conversations
The following two sentences are saying the same thing. Emily, my ...
Weegy: The following two sentences are saying the same thing. Emily, my daughter, is in sixth grade. Emily, my daughter ...
11/13/2017 12:26:02 PM| 3 Answers
3.85 x 10^-3
Weegy: (3.85 x 10) - 3 User: Give the number of significant figures in this number: 40.00 Weegy: number of ...
11/15/2017 9:18:26 AM| 2 Answers
Efficiency is
Weegy: Efficiency is the ability to do things well, successfully, and without extra waste of money, efforts, time etc. ...
11/7/2017 6:41:21 AM| 2 Answers
Permafrost (gelisol) is considered a hearty, robust soil type. ...
Weegy: Permafrost (gelisol) is considered a hearty, robust soil type. True. User: What is the purpose of strip ...
11/7/2017 6:50:38 AM| 2 Answers
Solve for x in the equation 6x = 42.
Weegy: 6x = 42
11/8/2017 11:42:29 AM| 2 Answers
4y + 228 = 352.
Weegy: 4y + 228 = 352
11/8/2017 11:59:04 AM| 2 Answers
Coefficient of 2xy^3
11/8/2017 12:00:24 PM| 2 Answers
Weegy Stuff
S
L
P
Points 166 [Total 291] Ratings 9 Comments 76 Invitations 0 Offline
S
P
P
L
P
P
Points 20 [Total 1658] Ratings 0 Comments 20 Invitations 0 Offline
S
P
L
P
Points 20 [Total 150] Ratings 0 Comments 20 Invitations 0 Offline
S
Points 1 [Total 1] Ratings 0 Comments 1 Invitations 0 Offline
S
Points 1 [Total 1] Ratings 0 Comments 1 Invitations 0 Offline
S
Points 1 [Total 1] Ratings 0 Comments 1 Invitations 0 Offline
S
Points 1 [Total 1] Ratings 0 Comments 1 Invitations 0 Offline
S
Points 1 [Total 1] Ratings 0 Comments 1 Invitations 0 Offline
* Excludes moderators and previous
winners (Include)
Home | Contact | Blog | About | Terms | Privacy | © Purple Inc.