Worksheet For Quadratics With A Coefficient Greater Than 1

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Homework 3
Amneris suggested the following factoring algorithm for quadratics with
2
a coefficient greater than 1 on the x
term as in the following example.
2
2
Suppose we want to factor (2x
3x
2). Multiply the coefficient of x
and
the constant term to get
4. We then look for factors of
4 that add to
3.
In this case, the factors are
4 and 1. We now write
(x
4)(x + 1).
Since a was 2, we check to see which constant term 2 factors. In this case, 2
is a factor of
4. We then divide
4 by 2 and put the 2 in front of the x in
(x + 1) to get
2
2x
3x
2 = (x
2)(2x + 1).
Often the process of finding the factors is done with the following graphic:
4
4
4
1
3
3
1. Explain why this method works. (Your explanation should include
a proof that the method works, not simply be a “heuristic” reason
why you might do this. However, it should also include a heuristic
explanation.)
There are two things to check. First, whether if the method
is successful that what arises is a solution, and second that if the
method is not successful then there is no factorization. We begin with
the first.
2
Suppose we are given the function ax
+ bx + c = 0 and s
and s
1
2
are integers such that s
s
= ac and s
+ s
= b. Moreover, assume
1
2
1
2
a s
. Then s
= a r
for some integer r
. Now the method gives the
1
1
1
1
factorization (x + r
)(ax + s
) for the polynomial. Multiplying this out
1
2
1

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