Question Video: Simplifying an Expression Involving Multiplication and Subtraction of Binomials and Trinomials | Nagwa Question Video: Simplifying an Expression Involving Multiplication and Subtraction of Binomials and Trinomials | Nagwa

# Question Video: Simplifying an Expression Involving Multiplication and Subtraction of Binomials and Trinomials Mathematics

A rectangle with dimensions (π₯ β π¦) and (π₯ + π¦ + 1) is cut out from a larger rectangle with dimensions (2π₯ + π¦ + 3) and (2π₯ + π¦).

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### Video Transcript

A rectangle with dimensions π₯ minus π¦ and π₯ plus π¦ plus one is cut out from a larger rectangle with dimensions two π₯ plus π¦ plus three and two π₯ plus π¦. Find a simplified expression for the shaded area.

We can find an expression for the shaded area by finding the area of the larger rectangle and subtracting the area of the smaller rectangle from the area of the larger rectangle. We recall the area of a rectangle is found by multiplying its length times its width. Letβs start by expanding and simplifying the expression for the area of the larger rectangle, two π₯ plus π¦ plus three times two π₯ plus π¦. We can do this by distributing the trinomial over each term in the binomial. So, we have two π₯ times two π₯ plus π¦ plus three plus π¦ times two π₯ plus π¦ plus three.

Then, we can distribute the monomials over the trinomials. Distributing two π₯ over two π₯ plus π¦ plus three gives four π₯ squared plus two π₯π¦ plus six π₯. And distributing π¦ over the same trinomial gives two π₯π¦ plus π¦ squared plus three π¦. We can add these expressions together and collect like terms to find the area of the larger rectangle, four π₯ squared plus four π₯π¦ plus π¦ squared plus six π₯ plus three π¦.

Now, letβs clear some space to find the area of the smaller rectangle. For the smaller rectangle, we have dimensions π₯ plus π¦ plus one and π₯ minus π¦. So, the expression for its area comes from the product of these two expressions. We can do this, once again, by distributing the trinomial over each term in the binomial. So, we have π₯ times π₯ plus π¦ plus one minus π¦ times π₯ plus π¦ plus one. Then, we can distribute the monomials over the trinomials. Distributing π₯ over π₯ plus π¦ plus one gives π₯ squared plus π₯π¦ plus π₯. And distributing negative π¦ over π₯ plus π¦ plus one gives negative π₯π¦ minus π¦ squared minus π¦. We can then add these expressions together and collect like terms to find the area of the smaller rectangle, resulting in the expression π₯ squared minus π¦ squared plus π₯ minus π¦.

Now we need to subtract the area of the smaller rectangle from the area of the larger rectangle to find the area of the shaded region. We must be careful on this step to subtract all the terms of the second polynomial, not just the first term. We can do this by distributing the negative through each term and then collecting the like terms. Thus, we subtract π₯ squared, add π¦ squared, subtract π₯, and add π¦ to the first expression. We can use a vertical method to stack the like terms, resulting in three π₯ squared plus four π₯π¦ plus two π¦ squared plus five π₯ plus four π¦. This is the simplified expression for the shaded region, which we found by subtracting the area of the smaller rectangle from the area of the larger rectangle.

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