Video: Expanding Two Linear Brackets

Expand the product (2π‘₯ + 1)(3π‘₯ βˆ’ 2).

03:02

Video Transcript

Expand the product two π‘₯ plus one, 3π‘₯ minus two.

In this question, the word product indicates that we’ll be multiplying our two binomials. We’re going to use two different methods to demonstrate the expansion of these binomials. The first is the FOIL method. And the second is the area or grid method. So starting with the FOIL method, this means that we take our first, then the outer, then the inner, and then the last terms and multiply. Looking at our binomials, the first terms are the two π‘₯ and the three π‘₯. So we multiply those, writing two π‘₯ times three π‘₯.

The outer terms will be our two π‘₯ and our negative two. So we add two π‘₯ times negative two. The inner terms will be our plus one and our three π‘₯. Therefore, we add our one times our three π‘₯. And finally, the last two terms will be the plus one and the negative two. So we add plus one times negative two. We may not always need to write this line of workings. But it can be helpful to illustrate our method. And now, we work out the products. Two π‘₯ times three π‘₯ is six π‘₯ squared. Two π‘₯ times negative two is negative or minus four π‘₯ plus one times three π‘₯ is equivalent to plus three π‘₯. And finally, plus one times negative two is negative two.

The next thing to do at this stage of the question is to simplify by collecting any like terms. Here, we have two terms in π‘₯. So we can write this as six π‘₯ squared minus π‘₯ minus two, because our negative four π‘₯ plus three π‘₯ is equivalent to a negative π‘₯. Let’s see how we would get the same answer using the grid method. To set up our grid, we put one binomial horizontally and one binomial vertically. The first binomial will give us two π‘₯ and one horizontally. And the second binomial of three π‘₯ minus two can be split into the terms three π‘₯ and negative two.

We then calculate the product of each row and column, starting with three π‘₯ times two π‘₯. This will give us six π‘₯ squared. Next, three π‘₯ times one is three π‘₯. In our second row, negative two times two π‘₯ is negative four π‘₯. And finally, negative two times one is negative two. We then simplify by adding all four terms and seeing if we can collect any like terms. We start with six π‘₯ squared. We then have a plus three π‘₯ and a negative four π‘₯, which simplifies to negative π‘₯ and then our final term of negative two. So using either of our methods would give us the answer of six π‘₯ squared minus π‘₯ minus two.

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