Question

Help!! Part 1:
In your own words, explain how a trinomial of the form 2x^2 + 13x + 15 can be turned into a four term polynomial suitable for factoring by grouping. Use complete sentences.
Part 2:
If you were an Algebra 1 instructor and were creating a test on factoring trinomials of the form ax^2 + bx + c, what do you think would be the easiest way to create a trinomial that can be factored? Provide one unique example.

Answer 1

ax^2 + bx + c is the trinomial
First, you multiply a nd c together.
Then you find two numbers whose product is the product ac and whose sum is b. Let's call these two numbers p and q.
Then you write the middle term, bx, as the sum px + qx, so the trinomial beocmes:
ax^2 + px + qx + c
Now you have 4 terms and you can factor by grouping.

Answer 2

Thanks so much!

Answer 3

@jim_thompson5910 can you help me with part 2~

Answer 4

how far did you get

Answer 5

I didn't get far at all, I just need help thinking of a simple one.

Answer 6

when a = 1, things are usually easiest

Answer 7

if a = 1, then all you need to do is look for two numbers that multiply to c, but also add to b

Answer 8

if a is not 1, then you can still come up with a trinomial that can be factored

just work backwards: start with something already factored and expand it out

Answer 9

example: (2x+1)(3x-5) is already factored

if you expand or FOIL that out, you will get something of the form ax^2 + bx + c

Answer 10

ask your instructor why you need to factor these types of trinomials by grouping in pairs...


try this for a trinomial your example

\[ax^2 + bx + c\]

multiply a and c 2 * 15 = 30

find the factors of a*c that add to b 10 + 3 = 13

then write

(ax + factor1)(ax + factor 2) (2x + 3)(2x + 10)
------------------------- ----------------
a 2

find the common factor 2(2x + 3)(x + 5)
----------------
2

cancel the common factor for the answer (2x + 3)(x + 5)

a simpler method that avoids

ask your instructor is he has seen this... Oh.... and it works every time.