Question Video: Expanding Algebraic Expressions Using Algebraic Identities | Nagwa Question Video: Expanding Algebraic Expressions Using Algebraic Identities | Nagwa

Question Video: Expanding Algebraic Expressions Using Algebraic Identities Mathematics

Expand (βˆ’π‘₯ + 2𝑦)Β².

02:16

Video Transcript

Expand negative π‘₯ plus two 𝑦 all squared.

In this question then, we have a binomial expression, negative π‘₯ plus two 𝑦. And we are squaring it. That means we’re multiplying this binomial by itself. So, we’re looking for the result of multiplying negative π‘₯ plus two 𝑦 by negative π‘₯ plus two 𝑦. There are numerous different methods that we can use. In this question, I’m going to choose to use the FOIL method. Now, we just need to be a little bit careful because one of the terms in our binomial is negative. And we don’t want to let this trip us up. We need to be really careful with the signs when we’re multiplying each pair of terms together.

So, F, remember, stands for firsts. We multiply the first term in each binomial together. That’s negative π‘₯ multiplied by negative π‘₯, which gives π‘₯ squared. Remember, a negative multiplied by a negative gives a positive. Then, the letter O stands for outers or outsides. We multiply the terms on the outside of our expansion. That’s the negative π‘₯ in the first binomial by the positive two 𝑦 in the second, giving negative two π‘₯𝑦. I stands for inners or inside. So, we multiply the terms in the center of our expansion. That’s the two 𝑦 in the first binomial and the negative π‘₯ in the second, giving another lot of negative two π‘₯𝑦. Finally, the letter L stands for lasts, so we multiply the last term in each binomial together. That’s positive two 𝑦 multiplied by positive two 𝑦, which is four 𝑦 squared.

So, after completing all four of our multiplications, we now have four terms in our expansion, π‘₯ squared minus two π‘₯𝑦 minus two π‘₯𝑦 plus four 𝑦 squared. Remember, there should always be two identical terms in the center of our expansion. And indeed, there are. We have negative two π‘₯𝑦 minus another lot of two π‘₯𝑦. We can therefore simplify our expansion by grouping like terms, and we have our final answer to the problem. The simplified expansion of negative π‘₯ plus two 𝑦 all squared is π‘₯ squared minus four π‘₯𝑦 plus four 𝑦 squared.

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