# Video: Pack 3 β’ Paper 2 β’ Question 17

Pack 3 β’ Paper 2 β’ Question 17

04:54

### 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 ππ₯ plus π over ππ₯ 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 weβll 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 weβre 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, itβs sensible to find common factors. To do this, we first factorize both the numerator and the denominator. Letβs 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 itβs two square numbers with a minus sign between them, itβs called the difference of two squares.

To factorize an expression, itβs the difference of two squares. For example, π squared minus π squared. Itβs 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, letβs 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. Weβll worry about the signs of these in a moment.

Now, letβs 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, weβre hoping that there are going to be some common factors. That means there should be one bracket in each of the expressions thatβs 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 weβve selected the numbers in each of our brackets, letβs 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, letβs 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.