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Answers
Answer:as real factors, they may often be found quite easily by using a.
simple device which forms the subject of this Appendix.
We know that
(ax +ex) (bx + {3) =abx2 + (af3 +bex)x +exf3
If we compare the right-hand side of this identity with
Ax2 +Bx+O we have
A=ab
B=af3+bex
0=exf3
We note that if A is positive, then a and b will be of the same
sign. If A is negative, a and b have different signs.
If C is positive, a and f3 have the same sign. If C is negative,
ex and f3 have different signs.
If B is negative, then at least one of the products af3 and ba must
be negative. This cannot be unless either a and f3 have opposite
signs, orb and ex have opposite signs. A positive B implies only that
at least one of the products is positive, and that therefore either
a and {3, orb and ex have the same signs.
All of this information is readily used by writing the result in the
form
abx2 + (af3 +ba)x +exf3
+ax~+ex
+b~ -...........+{3
(ax +ex)(bx + {3)
The term abx2 is the product of the terms on the left extremities
of the cross.
The term exf3 is the product of the terms on the right extremities
of the cross.
The term (af3 + bex)x is obtained by multiplying together the terms
521
522 APPENDIX 1
at the ends of the one diagonal, and adding to their product the
product of the terms at the ends of the other diagonal, i.e.,
(ax x {3) + (bx x IX.)
The rest of the method is best explained by taking some examples.
(i)
In this all the terms are positive. This means that a and b are of
the same sign and that IX and {3 are of the same sign.
Now since a has the same sign as b, and {3 the same sign as IX,
then a{3 has the same sign as b!X..
But the middle term is positive, which indicates that a{3 + biX is
positive. It follows that both a{3 and b!X. are positive, and that therefore a and {3 have the same signs. We thus have that a, b, IX and {3
are all of one sign. This is always the case when A, Band C in the
expression Ax2 + Bx + C are of the same sign.
Knowing that they are all of one sign, we shall assume a, b, IX and {3
to be positive. We shall later justify this.
Then we commence our solution by writing
6x2 +I9x+I0
+x~+
+~~+
(x+ )(x+
Now the possible factors of 6 are 6 and I, and 2 and 3. The possible factors of 10 are IO and I and 5 and 2. We now consider the
possible combination of these factors.
Try
This gives us
Ax2 +Bx+C
+6~<+I0
+x +I
Ax2 =abx = 6x2
It follows that
Bx= (a{3 +b1X)X= (6x xI)+ (x x IO) =I6x
c =IX.{3 = 10
Explanation:
Answer:
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