Video: Pack 2 β€’ Paper 3 β€’ Question 18

Pack 2 β€’ Paper 3 β€’ Question 18

03:17

Video Transcript

The following expression is an odd number when π‘₯ is an integer. π‘₯ squared multiplied by four π‘₯ minus one plus π‘₯ minus two all squared minus five. Prove this algebraically.

Before we start this question, it is worth remembering that any expression that can be written in the form two multiplied by 𝑛, where 𝑛 is an integer will represent an even number. This is because when we multiply an integer by two, our answer is always even.

Every even number is preceded by an odd number and it is also followed by an odd number. Therefore, any expression that can be written in the form two 𝑛 plus one or two 𝑛 minus one must represent an odd number, once again where 𝑛 is an integer.

Let’s now consider our expression. Our first step is to expand the brackets and simplify the expression. π‘₯ squared multiplied by four π‘₯ is four π‘₯ cubed and π‘₯ squared multiplied by negative one is negative π‘₯ squared. In order to expand π‘₯ minus two all squared, we need to write the bracket out twice and use the FOIL method.

Multiplying the first terms π‘₯ multiplied by π‘₯ gives us π‘₯ squared. Multiplying the outside terms gives us negative two π‘₯ as π‘₯ multiplied by negative two is negative two π‘₯. Multiplying the inside two terms also gives us negative two π‘₯. And finally, multiplying the last terms gives us positive four. Simplifying this by grouping our like terms gives us π‘₯ squared minus four π‘₯ plus four.

Our final step is to drop the minus five into the next line of our working. Grouping the like terms allows us to cancel the π‘₯ squareds as negative π‘₯ squared plus π‘₯ squared is equal to zero. We can also group positive four and negative five. This leaves us with four π‘₯ cubed minus four π‘₯ minus one. The first two terms are divisible by two. Therefore, we can factorize out a two. This gives us two multiplied by two π‘₯ cubed minus two π‘₯ minus one.

We were told in the question that π‘₯ is an integer. And if π‘₯ is an integer, this means that two π‘₯ cubed minus two π‘₯ will also be an integer. If we let two π‘₯ cubed minus two π‘₯ equal 𝑛, our expression can be written two 𝑛 minus one.

As the expression can be written in the form two 𝑛 minus one, we can say that the expression is an odd number when π‘₯ is an integer.

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