How do you factor trinomials of the type ax^2+bx+c?

Question

im in algebra, and we are learning about factoring expressions, and we're learning the type ax^2+bx+c. 

Answer 1

If ax²+bx+c is factorable (with integer coefficients), where a, b, and c are integers, then there are two integers whose product is ac and whose sum is b.  

 

For example, let's try to factor 10x²+11x-6.  We need to find two integers with product (10)(-6) = -60 and with sum 11.

 

The integers are 15 and -4.

 

Next, rewrite 11x as 15x-4x to get 10x²+15x-4x-6

 

               Factor by grouping:       5x(2x+3)-2(2x+3)

 

                                            =   (2x+3)(5x-2)

 

On the other hand, 3x²+5x-7 is not factorable (with integer coefficients) since there are not two integers with product (3)(-7) and sum equal to 5.  

                                                                  

Answer 2

if you are not able to find the factors then use this 

 

( -b +-  √ b² - 4 ac ) / 2a 

 

You can see the formula here also

http://www.biology.arizona.edu/biomath/tutorials/quadratic/roots.html

 

 

Answer 3

First, you need to figure out what a, b, and c represent.  I will tell you that a and b are coefficients of x² and x, respectively. c is the constant term.

 

Second, write down factor pairs for a and c.  Keep in mind that a is nonzero as a leading coefficient.

 

For example, x² - 3x - 4.  We have a = 1, b = -3, and c = -4.  The factor pairs for 1 is 1*1.  The factor pairs for 4 is 1 * -4 and 2 * -2.  Play with those numbers and you should end up with (x - 4)(x + 1).  Check to see if this leads to the original trinomial by using the FOIL method.

 

Hope this makes sense.