We run through our steps for factoring a quadratic trinomial.
When we multiply the first and last coefficients, (1)(4) = 4 so
we are looking for a pair of factors of 4 which add up to 5. There is
such a pair, 4 and 1. This gives us the coefficients of the O and I
terms when we recover the FOIL step.
At this point it is possible to recover the factored form. In this
case the greatest common factor of the F and O terms is a simple x.
Since the first term in the first binomial of the answer is one of the
First terms and is also one of the Outer terms, it would need to be
something that goes into both the F and O products. Let x be the
first term in the first binomial of the answer.
Once we have gotten this far the problem will truly crack if we
just fill in the terms in the parentheses that would make the problem
come out right if we removed the parentheses. The reason the
problem cracks at this point is because if we know that one of the
First terms is x, then we can figure that the other First term will
have to be what you would need to multiply an x by to get x
squared, namely, another x
At this point, we are halfway done. We have half of the terms
in the answer. We next turn our attention to the product of the Outer
terms. This is a -4x. This is the product of the first Outer term
which is the x in the first binomial of the answer times the second
term in the second binomial. We can conclude that the other Outer
term is a -4.
We should notice that we are doing fairly well at this point.
There is only one more term to fill in. There are two ways of looking
at this final missing term. It is one of the Inner terms. so it must be
what we would need to multiply the other Inner term of x by to get
-x. It is also one of the two Last terms, so it is what we would need to
multiply the other last term of -4 by to get the +4. The term which
will do both of those things is a -1.
This type of problem in which the first term in the quadratic
trinomial is a simple square term with a coefficient of an invisible 1
is a special, easier type of problem. The x squared comes from an x
and an x, and then the numbers multiply up to the last term and add
up to the coefficients of the middle term in the quadratic trinomial.
It is important to realize that this last mentioned technique will only
work on the special case where the coefficient of the first term is an
invisible 1. We shall see some examples later where this does not
happen. With the other problems which fall into this special category,
we omit the step by step discussion and go straight from the FOIL
step to the answer.
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