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When the dolphin is in the water. An equation for can be written by substituting for in the given equation. The solutions of the above equation represent the time when the dolphin leaves the water and the time when the dolphin is back in the water. The equation will be now solved. First, it will be rearranged, as it is usually preferred to have the variable terms on the left-hand side. Next, the greatest common factor (GCF) of the terms on the left-hand side will be factored out. The factors of and are listed below. The two terms have only one common factor Therefore, the GCF of and is The left-hand side of the equation is written in factored form and the right-hand side is equal to Therefore, this equation can be solved using the Zero Product Property. The solutions are and It is known that at the dolphin is still in the water. Therefore, it must be back in the water after seconds.

It is known that the area of the given rectangle is The area of a rectangle can be calculated by multiplying its dimensions — width and length. In this case, the width and the length of the rectangle are given as algebraic expressions. The product of and represents the area of the rectangle. Note that both dimensions of the rectangle are given in meters, so there is no need to change the units. The actual dimensions of the rectangle can be determined by solving the above equation. First, the equation will be rewritten in standard form by using the Distributive Property and Subtraction Property of Equality. The equation is written in the form The values of and can be identified. It can be seen that and Next, the quadratic trinomial on the left-hand side will be written in factored form following the general method. Note that and have a greatest common factor (GCF) of Next, the product of and will be found so that all pairs of integers whose product is equal to can be listed. The product and are both negative. Therefore, one of the factors has to be positive and one negative. Additionally, the absolute value of the negative factor has to be greater than the absolute value of the positive factor. The sum of the factors should be

Pair of Factors	Sum of Factors
and	$1+(\text{-} 280)=\text{-} 279\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$2+(\text{-} 140)=\text{-} 138\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$4+(\text{-} 70)=\text{-} 66\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$5+(\text{-} 56)=\text{-} 51\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$7+(\text{-} 40)=\text{-} 33\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$8+(\text{-} 35)=\text{-} 27\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$10+(\text{-} 28)=\text{-} 18\ {\color{#009600}{\bm{\Large{\checkmark }}}}$
and	$14+(\text{-} 20)=\text{-} 6\ {\color{#FF0000}{\bm{\Large{\times }}}}$

The pair of factors whose product is equal to and whose sum is is and Now, the linear term can be rewritten as a difference. Next, the GCF of the first two terms and the last two terms will be factored out. For more information on this process, read about finding the GCF. Note that for the factorization of the polynomial to work, -7 and not the GCF 7 has to be factored out from the last two terms. Finally, the common factor will be factored out. Since the left-hand side of the equation is written in factored form and the right-hand side is the equation can be solved using the Zero Product Property. Two solutions to the quadratic equation have been found. They will now be substituted into the expressions for the width and length of the rectangle.

	$2({\color{#0000FF}{\text{-} 1.25}})-1=\text{-} 3.5\ {\color{#FF0000}{\bm{\Large{\times }}}}$	$4({\color{#0000FF}{\text{-} 1.25}})-7=\text{-} 12\ {\color{#FF0000}{\bm{\Large{\times }}}}$
	$2({\color{#FF0000}{3.5}})-1=6\ {\color{#009600}{\bm{\Large{\checkmark }}}}$	$4({\color{#FF0000}{3.5}})-7=7\ {\color{#009600}{\bm{\Large{\checkmark }}}}$

It can be seen that results in a negative width and length, which is not logical for the dimensions of a rectangle. Therefore, the only valid solution is The width of the rectangle is meters and the length is meters.

To decide which girl is correct, the given equation will be solved by factoring. First, it will be rewritten in standard form. In this case and The greatest common factor (GCF) of these numbers is Now, all pairs of integer factors whose product is equal to will be listed. Since is positive and is negative, both factors have to be negative. Their sum will be also calculated in the table, it should be equal to

Pair of Factors	Sum of Factors
and	$\text{-} 1+(\text{-} 225)=\text{-} 226\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$\text{-} 3+(\text{-} 75)=\text{-} 78\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$\text{-} 5+(\text{-} 45)=\text{-} 50\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$\text{-} 9+(\text{-} 25)=\text{-} 34\ {\color{#FF0000}{\bm{\Large{\times }}}}$
and	$\text{-} 15+(\text{-} 15)=\text{-} 30\ {\color{#009600}{\bm{\Large{\checkmark }}}}$

It can be noted that the pair of factors whose product is equal to and whose sum is is and The linear term on the left-hand side will be rewritten as a difference and then the trinomial will be factored. Note that is a perfect square trinomial, so it could also be factored using the method described here. Since the left-hand side of the equation is written in factored form and the right-hand side is the equation can be solved using the Zero Product Property. Note that Equation (I) and (II) are the same. Therefore, they have the same solution. The given quadratic equation has one solution, so Magdalena is correct.