• Support PF! Buy your school textbooks, materials and every day products Here!

Factorising Quadractics

  • Thread starter Davio
  • Start date
  • #1
65
0
Hey guys, how would I factorise something like 2p^2+5p+3? Using the quadractic formula will give me P, howver not the factorised product ie. (2p+3) and (p +1). Previously I used trial and error, is there a mathematical technique to do this?
 

Answers and Replies

  • #2
1,752
1
Well if you want to use the Quadratic formula, you simply need to know what your coefficients are. I'm not going to type the quad. form. bc you can find that in your book but your coefficients are:

a = 2
b = 5
c = 3

From there, just plug and chug.
 
  • #3
1,752
1
If you want to factor it rather than using the quadratic formula, this is how you would go about it.

You need 2 numbers that give you the product of positive 3. Well you obviously only have 2 choices: 1,3 - Also, if you have a positive constant/third-term, then you need either 2 positive numbers or 2 negative numbers.

Next, pay attention to the sign of your middle term: It is positive, so that means your 2 numbers must be positive.

Lastly, you need two numbers that give you the product of your first time: 1,2

Multiply diagonally and add the two numbers, check if it gives you 5. If not, reverse the number for only one of them and try again. Then multiply downwards to see if you get your first and second terms. From there, write it as a product.

Example: a^2+2ab+b^2

[tex]a \ \ \ b[/tex]
[tex]a \ \ \ b[/tex]

ab+ab=2ab - Middle term checks out!

[tex](a+b)(a+b)[/tex]

[tex]2 \ \ \ 1[/tex]
[tex]1 \ \ \ 3[/tex]
 
Last edited:
  • #4
cristo
Staff Emeritus
Science Advisor
8,107
72
The way I would factorise that:
  • I look at the number in front of the p^2 term.. since it is 2, this implies that the factors must be (2p & m)(p & n) (where & denotes + or - to be decided)
  • The last sign is positive, thus the signs in the brackets are either both + or both -
  • The middle sign is positive, thus both signs in the brackets are +; we have (2p +m)(p +n)
  • Now, you want numbers (n,m) such that mxn=3 and 2n+m=5; the former expression gives us a choice of (3,1) and the latter fixes n=1, m=3
 
  • #5
65
0
Thanks for the replies guys
The way I would factorise that:
  • I look at the number in front of the p^2 term.. since it is 2, this implies that the factors must be (2p & m)(p & n) (where & denotes + or - to be decided)
  • The last sign is positive, thus the signs in the brackets are either both + or both -
  • The middle sign is positive, thus both signs in the brackets are +; we have (2p +m)(p +n)
  • Now, you want numbers (n,m) such that mxn=3 and 2n+m=5; the former expression gives us a choice of (3,1) and the latter fixes n=1, m=3
Ah I'll try that, that looks like what I'm looking for, *goes off and factorises things*
 
  • #6
cristo
Staff Emeritus
Science Advisor
8,107
72
No worries! I've moved to homework, since it fits in here better!
 
  • #7
238
0
I just wanted to add that you can use the quadratic formula to factorize as well, in case you didn't know that.

Once you have P the factors become (p-root1)(p-root2) or (p-root1)^2 if there is just one.

k
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,794
925
In this particular problem, the quadratic formula gives root1= -1, root2= -3/2. As kenewbie said, we can use (p- root1) and (p- root2) or, in this case, (p+1) and (p+ 3/2) as factors. Because 2p2+ 5p+ 3 has "2" multiplying p2, you will also need to multiply by 2: 2p2+ 5p+ 3= 2(p+1)(p+ 3/2)= (p+1)(2p+3).

If the the polynomial does NOT have factors with integer coefficients, you can still use the quadratic formula to factor. For example, the quadratic formula gives [itex]-1/2+ i\sqrt{3}/2[/itex] and [itex]-1/2- i\sqrt{3}/2[/itex] as roots of p2+ p+ 1= 0 so it can be factored as [itex](p+ 1/2- i\sqrt{3})(p+ 1/2+ i\sqrt{3})[/itex].
 

Related Threads for: Factorising Quadractics

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
941
Top