Factorization of Quadratic Expressions
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Key Questions

Factorization of a quadratic expression is the opposite of expansion, and is the process of putting the brackets back into the expression rather than taking them out.
To factorize a quadratic expression of the form
#ax^2+bx+c# you must find two numbers that add together to give the first coefficient of#x# and multiply to give the second coefficient of#x# .An example of this would be the equation
#x^2 + 5x + 6# , which factorizes to give the expression#(x+6)(x1)# Now, one might expect the solution to include the numbers 2 and 3, as these two numbers both add together to give 5 and multiply to give 6. However, as the signs differ in the factorized equation, then the solution to the equation must be
#(x+6)(x1)# , as#+6 1# gives#5# , and#6 times 1# yields a solution of 6.The equation can be checked by multiplying the solutions back into the equation to give the original quadratic of
#x^2 + 5x + 6# . 
Good question .
basically what are asking me to factor is :
#ax^2 + bx +c# here is an eg.
#4x^2 â€“ 15x + 9.# First you need to make column like this
SUM = COEFFICIENT OF x =( 15)
product = (COEFFICIENT of x^2#*# COEFFICIENT of constant)= 36
Now you have to find factors which have a sum of 15 and a product of 36THE FACTORS ARE (12), (3)
So now what we do is split the middle term in the equation and factor it out
#4x^2 â€“ 12xâ€“ 3x +9.#
#4x(x â€“ 3)â€“3 (x3).# #(4xâ€“3 )(x â€“ 3).# HENCE FACTORED
#f(4x^2 â€“ 15x + 9.) = (4xâ€“3 )(x â€“ 3)# 
Example 1
#x^2x6=(x+2)(x3)#
Example 2
#2x^29x5=(2x+1)(x5)#
Example 3
#x^29=(x+3)(x3)#
I hope that this was helpful.

Answer:
If both the roots satisfy the quadratic equation, then you have factorised it correctly.
Explanation:
I will try explaining with an example.
#x^2+4x12=0#
The roots are#2# and#6# .
Now simply put the values in the quadratic equation.First, I will try
#6#
#(6)^2+4(6)12=0#
#362412=0#
#cancel36cancel36=0#
#0=0#
which is true.Now, I will try
#2#
#2^2+4*212=0#
#4+812=0#
#cancel12cancel12=0#
#0=0#
which is also true.#:.# Our answer was correct.Hope you got it :)

Extension to factoring, when the trinomials do not factor into a square (it also works with squares).
Sumproductmethod
Say you have an expression like#x^2+15x+36#
Then you try to write#36# as the product of two numbers, and#15# as the sum (or difference) of the same two numbers. In this case (with both being positive) it's not so hard. You take the sum.You can write
#36=1*36=2*18=3*12=4*9=6*6#
Sums of these are#37,20,15,13,12# respectively
Differences are#35,16,9,5,0# respectively
#15=+3+12# will do. So the factoring becomes:
#(x+3)(x+12)#
Check your answer!#=x^2+12x+3x+36# It's a bit harder when one or two of the numbers are negative, let's take
#x^215x+36#
Same as the first, only now both factors are negative
#(x3)(x12)=x^212x3x+36=# the originalExtra
If the last number (#36# ) is negative, you will have to work with the difference of the factors. Check the next one yourself:
#x^2+5x36=(x+9)(x4)=?# And now try:
#x^25x36=?#
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Polynomials and Factoring

1Polynomials in Standard Form

2Addition and Subtraction of Polynomials

3Multiplication of Monomials by Polynomials

4Multiplication of Polynomials by Binomials

5Special Products of Polynomials

6Monomial Factors of Polynomials

7Zero Product Principle

8Factorization of Quadratic Expressions

9Factor Polynomials Using Special Products

10Factoring by Grouping

11Factoring Completely

12Probability of Compound Events