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Factoring Trinomials: I need a walkthrough of factoring the trinomials.

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2 Answers

What you want to do, is get the product of (x + a)(x + b) to equal x2 + 11x + 28.
 
If you multiply (x + a)(x + b), this equals x2 + ax + bx + ab = x2 + (a+b)x + ab.
Therefore. a+b = 11, and ab = 28.  What numbers multiply to equal 28 and add to equal 11?
 
Try 7 and 4.  therefore x2 + 11x + 28 = (x + a)(x + b) = (x + 4)(x + 7).
 
If this had been x2  + 5x - 6, I would ask what numbers multiply to equal 6 and subtract to equal 5.
 
Try 6 and 1.  Therefore x2 + 5x - 6 = (x + a)(x + b) = (x + 6)(x - 1).  Note that the product gives (a + b) as
 
-x + 6x = 5x, the middle term.  Always multiply be sure the factors are correct.
 
I hope this helps.
 
Kevin
As long as you don't have any coefficient for your 'a' term, then here is what you do:
 
Factor your 'c' term. Don't worry about positive or negative yet:
 
28 * 1
14 * 2
7 * 4
 
We need to decide which pair to use. Now we check the sign on the 'b' term. Our two factors need to add up to match our 'b' term exactly, and give us the exact 'c' term when multiplied. I this case, +7 * +4 = +28, and +7 + (+4) = +11.
7 * 4 matches our needs. Now what?
We set up our binomials.
We know that we will have to FOIL them to check our answer, so we can test FOIL at any time, as we create them. With that in mind, you should realize that the first term in both binomials is 'x', since F= x*x, or x2. So our binomials are started like this:
(x    )(x    )=0
Now we put the pair of numbers into the binomials:
(x+7)(x+4)=0
When we FOIL to check, the result is our original trinomial.
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The next step - Now we will need to know what 'x' equals. What can 'x' be to make the entire expression equal to zero? -7 and -4. When you plug them in for 'x', one or the other binomial equals zero, times the other one still equals zero.
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Actually what you did was multiply 'a' times 'c' and THEN factored to find the correct pair. 1 * 28 = 28. I mention this so that you know what to do when the 'a' coefficient is something other than 1.