Factoring Quadratic Trinomials
(when x2 coefficient is not 1)
Now we know what to do if a = 1, in ax2 + bx + c...
But what if it's not? What if it's 2? Or 3? Or -1? (*fake screaming*)
It's just a process that's a little more complicated, and sometimes called factoring by grouping (but I wouldn't take my word for it - I've heard factoring by grouping can mean all factoring polynomials).
First, make sure that you really know what a, b, and c stand for. Then read this:
2x2 + 12x + 16
|1. Find product ac.
2 * 16 = 32
2. Find two factors of ac that add up to b (call them y and z).
4 and 8
3. Rewrite the expression so it is ax2 + yx + zx + c
2x2 + 4x + 8x + 16
4. Put parentheses around the first two terms, and the last two.
(2x2 + 4x) + (8x + 16)
5. Factor the new polynomials (two terms inside the each of the sets of parentheses).
2x(x + 2) + 8(x + 2)
6. Now, the polynomials in the parentheses should be the same. Therefore, you can factor them out of the others.
(2x + 8)(x + 2)
So, ready for another one?
3x2 - 8x + 5
ac = 15
-3*-5 = 15, -3 + -5 = -8
3x2 - 3x - 5x +5
(3x2 - 3x) + (-5x + 5)
3x(x - 1) + -5(x - 1) **Note: You want to get the parentheses the same, so make sure your common factor does that.**
(3x - 5)(x - 1)
-x2 + 3x + 4
ac = -4
-1*4 = -4, -1 + 4 = 3
(-x2 - x) + (4x + 4)
-x(x + 1) + 4(x + 1)
(-x + 4)(x + 1)
to go back to factoring quadratic trinomials.
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