Worksheet For Quadratics With A Coefficient Greater Than 1 Page 3

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12
12
2
6
8
8
He then writes (x + 6)(x + 2) and then is lost because 2 divides both 2
and 6. What should the student do to factor with the given method?
Explain. (I do not want you to say that the student should use an-
other method, but rather explain how to use and/or adapt this method
(without factoring out the 2 first) to finish the problem.)
Since 2 6 and 2 2, we can actually use either term. If we
choose the 6 term, then the factorization will give
+
2x
8x + 6 = (x + 3)(2x + 2)
now factor 2 out
= (x + 3)2(x + 1)
= 2(x + 3)(x + 1).
Alternatively, if we choose the 2 term, we would have
+
2x
8x + 6 = (2x + 6)(x + 1)
now factor 2 out
= 2(x + 3)(x + 1).
In either case we end up with the same factorization, which is also what
comes out if we factor out a 2 at the beginning.
2
3. Can the method be adapted to factor 6x
+ 13x + 6? Why or why not?
Yes, using the expanded method above we can now factor,
looking for numbers that add to 13 and multiply to 36, we see that 4
and 9 fit the bill. Now we look for a factorization of 6 such that one
factor divides 4 and the other divides 6. Since 2 and 3 satisfy these
conditions, we have
4
9
2
6x
+ 13x + 6 =
3x +
2x +
2
3
= (3x + 2)(2x + 3).
3

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