Worksheet For Quadratics With A Coefficient Greater Than 1 Page 2

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we have:
2
(x + r
)(ax + s
) = ax
+ s
x + ar
x + r
s
1
2
2
1
1
2
2
= ax
+ (ar
+ s
)x + r
s
.
1
2
1
2
However ar
= s
so the coefficient of x above is s
+ s
= b as desired.
1
1
1
2
We also know that ac = s
s
= ar
s
. Canceling out the a, we have
1
2
1
2
c = r
s
as desired. Thus
1
2
2
(x + r
)(ax + s
) = ax
+ bx + c
1
2
as desired. Thus if the method works it gives a factorization.
2
What about the reverse direction? Suppose ax
+ bx + c = (d
x +
1
r
)(d
x + r
). In this case, the factorization method clearly does not
1
2
2
work if both d
and d
are different from 1. However, we can modify
1
2
the method slightly. To modify the method, we first find s
and s
with
1
2
the property that s
s
= ac and s
+ s
= b as before. Now look for
1
2
1
2
a factorization of a = d
d
that has the property that d
s
and d
s
1
2
1
2
2
1
(where d
s
means d
divides s
evenly). Now, we have s
= r
d
and
1
2
1
2
2
1
1
s
= r
d
and the factorization is (d
x + r
)(d
x + r
) as desired. To
1
2
2
1
1
2
2
see that this must give a factorization, note that if we start with the
factorization we have
a = d
d
1
2
b = d
r
+ d
r
1
2
2
1
c = r
r
.
1
2
Thus ac = d
d
r
r
and b = d
r
+ d
r
. Since s
= r
d
and s
= r
d
,
1
2
1
2
1
2
2
1
1
2
1
2
1
2
we have ac = s
s
and b = s
+ s
, so the method would produce an
1
2
1
2
answer if the original equation factors.
(Note that the first argument above can be used to show that the
adapted method produces solutions that work.)
2
2. A student in trying to factor 2x
+ 8x + 6 writes the following:
2

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