Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
5. Solving Polynomial Equations
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Exercise 64 Page 196

Recall the formula for the volume of a pyramid.

Base: 2feet* 2feet squared
Height: 3feet

Practice makes perfect
We will start finding the dimensions of the mold by recalling the formula for the volume of a pyramid. V=Bh/3 In this formula, B is the area of the base and h is the height of the pyramid. According to the information on the diagram, the base is an x* x square and the height is x+1. Let's use these expressions to write the volume in terms of x. V=x^2(x+1)/3 We are given that the volume is 4 cubic feet. Let's use this value to set up and simplify an equation for x.
x^2(x+1)/3=4
â–Ľ
Rearrange equation
x^2(x+1)=12
x^3+x^2=12
x^3+x^2-12=0
According to the Rational Root Theorem, the integer solutions of this equation are factors of the constant term -12. Since x represents a length, we are only interested in positive solutions. Let's list the positive factors of -12. 1, 2, 3, 4, 6, 12 We will substitute these into the left hand side of the equation to see whether any of these numbers is a solution.
x x^3+x^2-12 Value
1 1^3+ 1^2-12 - 10
2 2^3+ 2^2-12 0 âś“

We found that x=2 is a solution. Since the volume is increasing in x, there is only one x-value that gives the 4 cubic feet volume. This means that we do not need to look further, we found the only solution. Let's substitute x=2 into the measurements given on the diagram to find the dimensions of the mold. Base:& 2feet* 2feet squared Height:& 2+1=3feet