Alicia's family wants to save for a four-year college education for her.
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Area of a rectangle = W X L
Area of a rectangle = 5 X 2
Area of a rectangle = 10
Area of triangle 1= 1/2 X B X H
Area of triangle 1= 1/2 X 2 X 2
Area of triangle 1= 1/2 X 4
Area of triangle 1= 2
Area of triangle 2= 1/2 X B X H
Area of triangle 2= 1/2 X 7 X 4
Area of triangle 2= 1/2 X 28
Area of triangle 2= 14
Area of a rectangle + Area of triangle 1 + Area of triangle 2=
10 + 2 + 14 = 36
Area of a rectangle = 5 X 2
Area of a rectangle = 10
Area of triangle 1= 1/2 X B X H
Area of triangle 1= 1/2 X 2 X 2
Area of triangle 1= 1/2 X 4
Area of triangle 1= 2
Area of triangle 2= 1/2 X B X H
Area of triangle 2= 1/2 X 7 X 4
Area of triangle 2= 1/2 X 28
Area of triangle 2= 14
Area of a rectangle + Area of triangle 1 + Area of triangle 2=
10 + 2 + 14 = 36
If there are real roots to be found for this polynomial, the Rational Root Theorem and synthetic division are the best way to find them. I teach from a book that uses c and d for the possible roots of the polynomial. C is our constant, 2, and d is the leading coefficient, 1. The factors of 2 are +/- 1 and +/-2. The factors for 1 are +/-1 only. Meaning, in all, there are 4 possibilities as roots for this polynomial. But there are only 3 total (because our polynomial is a third degree), so we have to find the first one, at least, from our possibilities above. Let's try x = -1, factor form (x + 1). If there is no remainder when we do the synthetic division, then -1 is a root. Put -1 outside the "box" and the coefficients from the polynomial inside: -1 (1 2 -1 -2). Bring down the first coefficient of 1 and multiply it by the -1 outside to get -1. Put that -1 up under the 2 and add to get 1. Multiply 1 times the -1 to get -1 and put that -1 up under the -1 and add to get -2. -1 times -2 is 2, and -2 + 2 = 0. So we have our first root of (x+1). The numbers we get when we do the addition along the way are the coefficients of our new polynomial, the depressed polynomial (NOT a sad one cuz it hates math, but a new polynomial that is one degree less than that of which we started!). The new polynomial is 
. That can also be factored to find the remaining 2 roots. Use standard factoring to find that the other 2 solutions are (x+2) and (x-1). Our solutions then are x = -2, -1, 1, choice B from above.