How can I factorise quadratics using the quadratic formula?

  • Thread starter Davio
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In summary, to factorize a polynomial, you can either use the quadratic formula to find the roots and then write the factors as (p- root1)(p- root2) or (p- root1)^2, or you can use the method of finding two numbers that multiply to the constant term and add to the coefficient of the middle term. The factors can then be written as (p+ a)(p+ b) where a and b are the two numbers.
  • #1
Davio
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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?
 
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  • #2
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
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
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
Thanks for the replies guys
cristo said:
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
No worries! I've moved to homework, since it fits in here better!
 
  • #7
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
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].
 

1. What is factorising quadratics?

Factorising quadratics is a method used to break down quadratic equations into simpler expressions, known as factors. This is done by finding two numbers that multiply to equal the constant term and add to equal the coefficient of the middle term in the quadratic equation.

2. Why do we factorise quadratics?

Factorising quadratics helps us solve equations, find the x-intercepts or roots of a graph, and simplify complex expressions. It also helps us understand the relationship between the factors and the graph of the quadratic equation.

3. What are the steps for factorising quadratics?

The general steps for factorising quadratics are:
1. Write the quadratic equation in the form ax^2 + bx + c
2. Find two numbers that multiply to equal c and add to equal b
3. Rewrite the middle term using these two numbers
4. Factor the expression by grouping
5. Check your answer by expanding the factors and simplifying.

4. Can all quadratic equations be factorised?

No, not all quadratic equations can be factorised. Some equations have complex solutions or do not have rational solutions, making it impossible to find two numbers that can be factored. However, the quadratic formula can be used to solve these types of equations.

5. How does factorising quadratics relate to real-life applications?

Factorising quadratics has various real-life applications, such as in physics, economics, and engineering. For example, it can be used to determine the maximum or minimum value of a quadratic function, which is important in optimizing production costs or finding the maximum height of a projectile. It is also used in cryptography and data encryption to break down complex codes.

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