# Video: Finding the Length of a Rectangle given Its Area and Width by Dividing Two Polynomials Then Evaluating Its Perimeter

A rectangle has an area of (𝑦³ + 2𝑦² + 5𝑦 + 10) cm² and a width of (𝑦 + 2) cm. Find its length in terms of 𝑦 and its perimeter when 𝑦 = 4.

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### Video Transcript

A rectangle has an area of 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10 square centimeters and a width of 𝑦 plus two centimeters. Find its length in terms of 𝑦 and its perimeter when 𝑦 equals four.

Well the area of a rectangle is calculated by multiplying the length by the width. If we let the length of the rectangle equal 𝑥 centimeters, then 𝑥 multiplied by 𝑦 plus two must be equal to 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10. Dividing both sides by 𝑦 plus two allows us to calculate the length by dividing 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10 by 𝑦 plus two.

We are going to now use long division to work out the length of the rectangle. In order to divide 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10, a cubic, by 𝑦 plus two, a linear expression, we need to divide, multiply, and subtract. Firstly, we need to divide 𝑦 cubed by 𝑦. This gives us 𝑦 squared. Secondly, we need to multiply this 𝑦 squared by 𝑦 plus two to work out our remainder. 𝑦 squared multiplied by 𝑦 is 𝑦 cubed, and 𝑦 squared multiplied by two is two 𝑦 squared.

Our next step is to subtract those two lines. Well 𝑦 cubed minus 𝑦 cubed is zero. And in this case, two 𝑦 squared minus two 𝑦 squared is also zero. Therefore, there is no remainder. This means that we do not need to divide the 𝑦 squared term by 𝑦. So we can drop down the five 𝑦 plus 10. We now repeat the process by dividing five 𝑦 by 𝑦. Well, as five 𝑦 divided by 𝑦 is equal to five, we can put five on the answer line. Multiplying this five by 𝑦 plus two gives us five 𝑦 plus 10 as five multiplied by 𝑦 is five 𝑦 and five multiplied by two is 10.

When we subtract these two terms again, we can see that once again there is no remainder as five 𝑦 minus five 𝑦 is zero and 10 minus 10 is also zero. Therefore, we can say that 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10 divided by 𝑦 plus two is equal to 𝑦 squared plus five. Therefore, the length of the rectangle is 𝑦 squared plus five. At the end of this video, I will show you an alternative method for calculating the length in terms of 𝑦. However, we will now focus on the second part of the question: finding the perimeter.

As we now know both of width and the length of the rectangle, we can substitute in the value 𝑦 equals four to calculate its perimeter. The width of the rectangle is four plus two, which is six centimeters. And the length is four squared plus five, which is 21 centimeters, as four squared is equal to 16 and 16 plus five equals 21. To calculate the perimeter of any shape, we need to work out the distance around the outside of the shape. In this case, we need to add 21, six, 21, and six.

Adding these four numbers gives us 54 centimeters. Therefore, the perimeter of the rectangle is 54 centimeters. Therefore, if a rectangle has an area of 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10 and a width of 𝑦 plus two, its length is 𝑦 squared plus five. And the perimeter when 𝑦 equals four is 54 centimeters. We will now look at an alternative method for finding an expression for the length in terms of 𝑦.

As the area of a rectangle is calculated by multiplying the width by the length, we know in this case that 𝑦 plus two multiplied by something is equal to 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10. As 𝑦 plus two is a linear equation and 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10 is a cubic equation, the missing term, in this case the length, must be a quadratic expression as a linear expression multiplied by a quadratic expression gives us a cubic expression.

Considering the general quadratic expression 𝑎𝑦 squared plus 𝑏𝑦 plus 𝑐, we can see that 𝑦 plus two multiplied by this quadratic expression gives us 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10. Expanding on multiplying out the brackets or parentheses gives us 𝑎𝑦 cubed plus 𝑏𝑦 squared plus 𝑐𝑦 when we multiply the quadratic by 𝑦 and two 𝑎𝑦 squared plus two 𝑏𝑦 plus two 𝑐 when we multiply the quadratic by two. We know that this must be equal to the cubic expression for our area.

Comparing coefficients gives us four equations. For the 𝑦 cubed terms, 𝑎 equals one. For the 𝑦 squared terms, 𝑏 plus two 𝑎 equals two. For the 𝑦 terms, 𝑐 plus two 𝑏 equals five. And for the free terms, two 𝑐 equals 10. Well the first equation tells us that 𝑎 is equal to one. And dividing the fourth equation by two tells us that 𝑐 is equal to five. We can now substitute either these two values into the second or third equation to work out the value of 𝑏.

Well as five plus two 𝑏 is equal to five, 𝑏 must be equal to zero. Substituting these back into our quadratic expression gives us one 𝑦 squared plus nought 𝑦 plus five. This can be simplified to give us the expression for the length of 𝑦 squared plus five. These are just a couple of different ways of finding the length in terms of 𝑦.