7. Factor

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.

F O I L

*x*^{2} - 4*x* - *x* + 4
^{
}

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 -4*x*. 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 that 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.