Quadratic Functions, Dividing Polynomials, and Zeros of Polynomials

Question

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Quadratic Functions:

1. You have 150 yards of fencing to enclose a rectangular region. One side of the rectangle does not need fencing. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

 

Complete the following steps to solve the above problem:

a. Write the equation for the area of the rectangular region:

A =

b. Write the equation for the fencing required:

150 =

c. Solve the equation for fencing for y.

d. Substitute the result of step c) into the area equation to obtain A as function of x.

e. Write the function in the form of . F(x)= ax^2+bx+c.

f. Calculate -b/2a . If a < 0, the function has a maximum at this value.

This means that the area inside the fencing is maximized when = ?

g. Find the length of side y.

h. Find the maximum area.

Synthetic Division:

2. Find all the roots/zeros of 2x^3-3x^2-45x-54 =0

Complete the following steps to solve the above problem:

a. Use synthetic division to divide 2x^3-3x^2-45x-54 by x+3

[Hint: Delete the question mark to input each answer.]

−3      2    −3    −45    −54

   +            ?       ?       ​?  

2       ?       ?       0

b. Our answer gives us the following quadratic equation: . Divide this by ( − 6) using synthetic division:

6      2    −9    −18

  +           ?        ?

       2     ?         ?

Remainder Theorem:

3. Use the remainder theorem to find for the equation .

−4      2    −3    −45    −54

   +             ?       ?        ?  

2      ?       ?         ?

Answer 1

1. The maximum area is when the rectangle is a square. So the three fenced sides are equal and have length 150/3=50 yd.

Let's look at the logic: the sides of the rectangle are L and W so the area, A, is LW and the perimeter, P=2L+2W. L=(P-2W)/2 and A=W(P-2W)/2=(1/2)(WP-2W^2). The maximum value of A is obtained by differentiating the quadratic with respect to W: (1/2)(P-4W)=0 at a turning point, so W=P/4 and L=(P-P/2)/2=P/4. Therefore L=W=P/4. The enclosed area is therefore a square, and we know that the three sides plus another unfenced length, L, make up the perimeter P=4L=150+L, so 3L=150 and L=50 yd.

Alternative solution avoiding calculus:

a. A=LW where L=length of rectangular area, W=width.

b. 150=2L+W.

c. W=150-2L. But the true perimeter of the area is P=2L+2W, so W=(1/2)(P-2L).

d. A=L(P-2L)/2=(LP-2L^2)/2=LP/2-L^2.

e. f(L)=-L^2+LP/2-A in standard quadratic form f(x)=ax^2+bx+c, where a=-1, b=P/2, c=-A and x=L.

f. -b/2a=(-P/2)/(-2)=P/4. This is the vertex of f(L), and L=P/4 is the value for the maximum area.

g. From (c) W=(1/2)(P-2L)=P/4, so W=L=P/4. This makes the area a square of area P^2/16.

From (b) 150=2L+W=P/2+P/4=3P/4, making P=4/3*150=200, so W=L=200/4=50 yd.

h. Maximum area is 50^2=2500 sq yd.

2. Synthetic division (1)

-3 | 2 -3 -45 -54

......2 -6..27...54

......2 -9 -18 | 0

3. Synthetic division (2)

6 | 2..-9 -18

.....2 12..18

.....2...3..| 0

This quotient is 2x+3, representing the third factor of the polynomial.

4. Synthetic division (3)

-4 | 2 -3 -45..-54

......2 -8..44.....4

......2 -11 -1 | -50

The Remainder Theorem tells us that substituting x=-4 into the polynomial gives us the same remainder. The substitution gives: -128-48+180-54=-50.

The zeroes are -3, 6 and -3/2, because synthetic division (2) tells us that 2x+3 is a factor.