In Exercises 53–54, write a polynomial that represents the length...

Question

In Exercises 53–54, write a polynomial that represents the length of each rectangle. Transcription: The area of the rectangle is 0.5x^3 - 0.3x^2 + 0.22x + 0.06 square units and its width is x + 0.2 units

Rectangle with width x + 0.2 units and area 0.5x^3 - 0.3x^2 + 0.22x + 0.06 square units.

Accepted Answer

Welcome back. I am so glad you're here. We have a rectangle and we're told that the rectangle has an area of 0.7 X cubed minus 0.695 X squared minus 0.2 X plus 0.15 interesting area. But okay. We are told that the length of the rectangle is X plus 0. units and we are asked to find its width. Okay, well we know that the area of a rectangle is equal to the length times the width. And since we know the length we can substitute that in for our length. So we'll have X plus 0. times W is equal to this big long polynomial here. 0.7 X cubed minus 0.695, X squared minus 0.2 X plus 0.15. Okay well we've got ourselves an equation and now we can solve for w let's divide both sides by X plus 0.15. And on the right hand side that's not too bad because that X plus 0.15 just cancels out and all we're left with is a W. Great on the left. We just have to simplify so how do we simplify this? Well we just need to do the division and there are a lot of different ways to do the division but I suggest using synthetic division. We look at our X plus 0.15 as X minus C. And then solving for C. Which will be used in our synthetic division. We subtract X from both sides. And we have 0.15 equals negative c divide both sides by -1. And we get C equals negative 0.15. So we put that in over here for our synthetic division and then we just bring down all of our coefficients and our constant for the synthetic division. So in front of this X cubed is that 0.7 and then continuing down the line with the coefficients and the constant -0.695, -0.02 and 0.015. Great. We recall from previous lessons the steps for synthetic division. The first step is the one that only happens the first time is to bring down that first term. So bring down this 0.7. So we've got a 0.7 down here and now the two steps that will be repeated over and over are 21 multiply So we take zero point or negative 0.15 and multiply it by 0.7. Then we put it up here and we get negative 0.105. And the second step is to add we're going to add negative 0.695 plus negative 0.105. And we get negative 0.8. Now we repeat we multiply again negative 0.15 times negative 0.8 and we will get a positive 0.12. Now we go back to adding we add these two negative 0.2 plus 0.12 gives us 0.1. Now we multiply again negative 0.15 times 0.1 gives us a negative 0.015 And we add 0.015 plus negative 0.015 and we get zero so there is no remainder. We recall from previous lessons that when we're doing synthetic division, this result we get down here is our new polynomial expression and we're going to use that 0.7 and it's going to go in front of a variable that is one degree lower than the one we started off with. So 0.7. The first term here was in front of an X. Cute. Now it goes in front of an X squared and then we have minus 0.8 going down one degree from the previous variables exponents, this will just be X. And then this last one here is going to be our constant. So plus 0.1 and that is the polynomial that describes our width. We look at our answer choices and this matches with answer choice. B. Well done. We'll catch you on the next one