Question

i dont understand factoring trinomials of the form ax^2+bx+c

Answer 1

Would working a few examples help?

Answer 2

yes pleas i wasn't there for the whoe clas tday and he still av me home work di really dnt know what im doing

Answer 3

If you have a few to specifically work on, that'd be great :) if not, we can look at a pretty simple one to start off

Answer 4

7x^2+17x+6

5x^2-18x+16

12x^2-40x+25

Answer 5

Cool, let's look at the first one.
\[7x^2 + 17x + 6\]We can do this by trial and error.

Let's break up the 7x^2 term into its factors. Since 7 is prime, this is pretty easy:\[7x, x\]
And we'll do the same for 6.\[1, 2, 3, 6\]
Now we're going to put these together, using different sets of factors. We can automatically fill in the x terms, since there's only one set of factors:
\[(7x + ____)(x + ____)\]
Now we're left with the factors of 6 to fill in the gaps. Remember that they have to multiply together to get 6, and add to get 17. This won't happen by itself- but we have that 7 to help it out. Let's try plugging in 1 and 6 and see what happens
\[(7x+6)(x+1)\]
Now, if we foil that out, we end up with\[7x^2+7x+6x+6 = 7x^2+11x+6\] So not quite.
How about 2 and 3?
\[(7x + 3)(x + 2)\]
Which gives us
\[7x^2+14x+3x+6 = 7x^2+17x+6\]
A-ha! That one checks out, and we're done! Make sense?

Answer 6

okay i got that one right next

Answer 7

The next two have a "minus plus" structure, so when you break it down, you'll end up with "minus minus", just like normal factoring... meaning we'll have two negative factors of your last term that will multiply to become positive and add together to result in the middle negative term.

Answer 8

okay

Answer 9

what do we do now

Answer 10

The first term of \[5x^2 - 18x + 16\] is prime again, so we'll plug it into the parentheses:\[(5x - ____)(x - ____)\] With me so far?

Answer 11

yea

Answer 12

Alright, let's look at some factors of 16\[1, 2, 4, 8, 16\]
4 is kind of a nice number, so let's start with it:\[(5x-4)(x-4)\] Which gives us \[5x^2-20x-4x+16 = 5x^2-24x+16\]
Nope... too big. Try out the other factor pairs (1 and 16 & 2 and 8) and see which one gives us the right factorization!