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.
26,922,469 questions answered
*
Get answers from Weegy and a team of really smart lives experts.
Popular Conversations
6 + (2 + 3) × 5?
Weegy: 2*6+4*5, 12 + 20 = 32 User: 2(5 × 3) + 2(5 × 8) + 2(3 × 8) Weegy: 3/8+2/5 = 15/40+16/40 = 31/40 User: ...
5/22/2018 9:36:32 AM| 2 Answers
Strike-slip fault
Weegy: In a strike-slip fault, the rocks on either side of the fault slip past each other sideways. TRUE.
5/17/2018 3:51:47 PM| 2 Answers
After the skin is cut and a capillary wall breaks, what's the first ...
Weegy: After the skin is cut and a capillary wall breaks, the first step in working to form a clot is - Platelets clump ...
5/23/2018 5:14:16 PM| 2 Answers
Which is not a responsibility of the president
5/18/2018 10:17:31 PM| 2 Answers
What does velocity mean
5/23/2018 7:13:12 PM| 2 Answers
Adverb clauses modify __________.
Weegy: An adverb clause is a dependent clause that acts as an adverb. User: Which sentence correctly forms the ...
5/19/2018 2:53:57 PM| 2 Answers
S
L
Points 226 [Total 237] Ratings 0 Comments 226 Invitations 0 Offline
S
R
L
R
P
R
Points 216 [Total 658] Ratings 0 Comments 26 Invitations 19 Offline
S
L
Points 141 [Total 141] Ratings 0 Comments 21 Invitations 12 Offline
S
1
L
L
P
R
P
L
P
P
R
P
R
P
R
P
Points 124 [Total 13209] Ratings 0 Comments 114 Invitations 1 Offline
S
Points 30 [Total 30] Ratings 0 Comments 0 Invitations 3 Offline
S
Points 14 [Total 14] Ratings 0 Comments 14 Invitations 0 Offline
S
Points 13 [Total 13] Ratings 0 Comments 3 Invitations 1 Offline
S
Points 13 [Total 13] Ratings 0 Comments 13 Invitations 0 Offline
S
Points 11 [Total 11] Ratings 1 Comments 1 Invitations 0 Offline
S
L
1
R
Points 11 [Total 1416] Ratings 0 Comments 1 Invitations 1 Offline
* Excludes moderators and previous
winners (Include)
Home | Contact | Blog | About | Terms | Privacy | © Purple Inc.