Exercise 51 Page 70

Let's begin by writing the other dimensions in terms of the width, w. height= w+9 length=w-4 We can create a polynomial for the volume of the prism in terms of w, as the product of these dimensions. w ( w+9)(w-4)Next, we simplify the expression.

w ( w+9)(w-4)

Distr

Distribute ( w+9)

w (w( w+9)-4( w+9) )

▼

Distribute w and -4

Distr

Distribute w

w (w^2+9w-4( w+9) )

Distr

Distribute -4

w (w^2+9w-4w-36 )

AddSubTerms

Add and subtract terms

w (w^2+5w-36 )

Distr

Distribute w

w^3+5w^2-36w

Next, we create an equation with the expression equal to the volume, 180, and solve for the width.

w^3+5w^2-36w=180

SubEqn

LHS-180=RHS-180

w^3+5w^2-36w-180=0

We can split the equation into two parts and factor out the common factors. ( w^3+5w^2)+(-36w- 180) ⇕ w^2(w+5)-36(w+5) ⇕ (w^2-36)(w+5) Notice that (w^2-36) is a difference of squares meaning it can be factored as follows. (w^2-36)(w+5) ⇕ (w+6)(w-6)(w+5) Then, we can set each factor equal to 0 and solve for w. rcl w+6=0 & ⇒ & w= -6 w-6=0 & ⇒ & w=6 w+5=0 & ⇒ & w=-5 We can identify that the only positive solution is w=6. Therefore, the width is 6ft and we can use initial expressions we created for length and height to find the other dimensions.

	Expression	Substitution	Dimension
Width	--	--	w=6
Height	h= w+9	h= 6+9	h=15
Length	l= w-4	l= 6-4	l=2