### Video Transcript

The expression 12π₯ squared minus
five π₯ minus two to the power of negative one multiplied by nine π₯ squared minus
four can be simplified to the form ππ₯ plus π over ππ₯ plus π, where π, π, π,
and π are integers. Find π, π, π, and π.

These questions are fairly common
in higher exam papers. However, this one looks very
different from the ones we usually come across. We can rewrite it slightly so that
it looks similar to those we know by considering what the power of the first bracket
means.

A negative power gives us the
reciprocal of an expression or number. Remember the reciprocal of π₯ is
just found by dividing one by π₯. That means π₯ to the power of
negative π¦ is equal to one over π₯ to the power of π¦. For instance, two to the power of
negative four is equal to one over two to the power of four or one over 16. Seven to the power of negative two
is one over seven squared which is one over 49.

In this case, 12π₯ squared minus
five π₯ minus two to the power of negative one is the same as one over 12π₯ squared
minus five π₯ minus two to the power of one. Anything to the power of one is
itself. So weβll disregard that power at
this point. That means the expression in our
question becomes one over 12π₯ squared minus five π₯ minus two multiplied by nine π₯
squared minus four.

Remember when weβre multiplying
fractions, we can write the second fraction as nine π₯ squared minus four over one
without changing its value. Finally, we can multiply these two
expressions. Finally, multiplying the numerator
of the first fraction by the numerator of the second gives us nine π₯ squared minus
four. Similarly, multiplying the
denominators gives us 12π₯ squared minus five π₯ minus two.

Now, this is where the expression
starts to look recognizable. Whenever we have an algebraic
fraction which needs simplifying, itβs sensible to find common factors. To do this, we first factorize both
the numerator and the denominator. Letβs start with the numerator. We know this factorizes into two
brackets since there are no common factors in each part of the expression. In fact, since itβs two square
numbers with a minus sign between them, itβs called the difference of two
squares.

To factorize an expression, itβs
the difference of two squares. For example, π squared minus π
squared. Itβs π plus π multiplied by π
minus π. Nine π₯ squared minus four
factorizes to three π₯ plus two multiplied by three π₯ minus two. Now, letβs factorize the
denominator. Since two is a prime number, the
only possible numbers that could be in the final part of each bracket are one and
two. Weβll worry about the signs of
these in a moment.

Now, letβs consider what the first
term in each bracket could be. The factor pairs of 12π₯ squared
are 12π₯ and π₯, six π₯ and two π₯, or four π₯ and three π₯. There is a little trick here. If we look at the expression nine
π₯ squared minus four, that factorizes to three π₯ plus two multiplied by three π₯
minus two. When we simplify our fraction,
weβre hoping that there are going to be some common factors. That means there should be one
bracket in each of the expressions thatβs the same.

One of the brackets in the
factorized form of 12π₯ squared minus five π₯ minus two must have both a three π₯
and a two in it then. It also means that the first
bracket must have that other factor pair of 12π₯ squared. It must have four π₯ in it. Once weβve selected the numbers in
each of our brackets, letβs expand it back out using the FOIL method to check our
signs.

Four π₯ multiplied by three π₯ is
12π₯ squared. Four π₯ multiplied by two is eight
π₯. One multiplied by three π₯ is three
π₯. And one multiplied by two is
two. In fact, we know it needs to be
negative two, which tells us that one of the signs for either one or two is going to
be negative. To make negative five π₯, we need
negative eight π₯ plus three π₯. That means that the sign in the
first bracket is a positive and the sign in the second bracket is a negative.

Now that we fully factorized both
the numerator and the denominator, letβs put them back into our algebraic
fraction. Notice that both the numerator and
the denominator share the common factor of three π₯ minus two. This means we can divide through by
three π₯ minus two. Our fraction simplifies to three π₯
plus two over four π₯ plus one.

Remember the question was asking us
to find the value of π, π, π, and π. π is three, π is two, π is four,
and π is one.