Question 199746
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I'm unsure how to measure the area of a 'sale'.  Perhaps it is the area of the floor surface in a store where a sale is being held.  However, I suspect you meant 'sail' since that is the word you used later in the problem, and I will proceed on that assumption.  Furthermore, I have no idea what the 'hignt' or 'lenght' of a sail (or a sale for that matter) is.  I'm going out on a limb here and will assume you meant 'height' and 'length'.  The point of this part of my lesson is that if I am going to take the time to help you, you are obligated to take the time to communicate correctly and completely.  Enough said about that.


If


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x) = x^2 + 5x + 6]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(5) =(5)^2 + 5(5) + 6]


The arithmetic is left as an exercise for the student.


The area of a triangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A = \frac{bh}{2}]


Therefore, the measure of the base (your length), given the area and the height, is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b = \frac{2A}{h}]


Since the area as a function of *[tex \Large x] is the function *[tex \Large A(x) ], the measure of the base, which is equivalent to the length that is the answer to this problem can be determined by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b = \frac{2(x^2 + 5x + 6)}{x + 3}]


There are two ways to proceed from here.  One, you could use polynomial long division to obtain the quotient *[tex \Large \frac{x^2 + 5x + 6}{x + 3}].  The other way, and much simpler, is to simply factor the polynomial in the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b = \frac{2(x + 2)(x + 3)}{x + 3}]


Then eliminate the factor common to both numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b = 2(x + 2) = 2x + 4]


Done.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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