Exercise 16 Page 381

Practice makes perfect

a Let h be the height of the given prism. We will rewrite length l and width w using h.

l=h-1 w=h+3 Recall that the volume of the rectangular prism is the product of all its dimensions. Since we know that the volume equals 864 cubic centimeters we can write that the product of length, width and height is equal to 864.

l * w * h=864

SubstituteII

l= h-1, w= h+3

( h-1)( h+3)h=864

▼

Simplify

Distr

Distribute (h-1)

[h(h-1)+3(h-1)]h=864

Distr

Distribute h

[h^2-h+3(h-1)]h=864

Distr

Distribute 3

[h^2-h+3h-3]h=864

AddTerms

Add terms

[h^2+2h-3]h=864

Distr

Distribute h

h^3+2h^2-3h=864

SubEqn

LHS-864=RHS-864

h^3+2h^2-3h-864=0

b Let's recall the Fundamental Theorem of Algebra.

Ifh(x) is a polynomial of degreen≥ 1, thenh(x)=0 has exactlyn roots. Those roots can be either real or imaginary. Both real and imaginary numbers are complex numbers. The polynomial in the given equation has a degree of 3. Thus, the equation has exactly three complex roots.

c Recall that any rational root for h(x)=0 has the form ± p q, where p is a factor of the constant term and q is a factor of the leading coefficient. Since q is equal to 1 we need to consider only factors of p. Let's substitute the positive factors of -864 into the polynomial until we find the first zero.

Positive factors of -864	h^3+2h^2-3h-864? =0	Is it a zero?
1	1^3+2( 1)^2-3( 1)-864? =0	-864≠ 0 *
2	2^3+2( 2)^2-3( 2)-864? =0	-854≠ 0 *
...	...	...
8	8^3+2( 8)^2-3( 8)-864? =0	-248≠ 0 *
9	9^3+2( 9)^2-3( 9)-864? =0	0= 0 ✓

We see above that a zero occurs at h=9. Let's use synthetic division to depress the polynomial and see if there are some other possible solutions.

rl IR-0.15cm r 9 & |rr 1 & 2 & -3 & -864

Bring down the first coefficient

rl IR-0.15cm r 9 & |rr 1 & 2 & -3 & -864 & c 1 & & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 9 & |rr 1 & 2 & -3 & -864 & 9 & & & c 1 & & &

Add down

rl IR-0.15cm r 9 & |rr 1 & 2 & -3 & -864 & 9 & & & c 1 & 11 & &

▼

Repeat the process for all the coefficients

Multiply the coefficient by the divisor

rl IR-0.15cm r 9 & |rr 1 & 2 & -3 & -864 & 9 & 99 & & & c 1 & 11 & &

Add down

rl IR-0.15cm r 9 & |rr 1 & 2 & -3 & -864 & 9 & 99 & & c 1 & 11 & 96 &

Multiply the coefficient by the divisor

rl IR-0.15cm r 9 & |rr 1 & 2 & -3 & -864 & 9 & 99 & 864 & c 1 & 11 & 96 &

Add down

rl IR-0.15cm r 9 & |rr 1 & 2 & -3 & -864 & 9 & 99 & 864 & c 1 & 11 & 96 & 0

The above are the coefficients of the depressed polynomial. h^3+2h^2-3h-864=0 ⇕ (h-9) ( h^2+11h+96 )=0 Now the depressed is a quadratic polynomial, where a= 1, b= 11, and c= 96. This means we can use the Quadratic Formula to find its zeros.

h=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

h=- 11±sqrt(( 11)^2-4( 1)( 96))/2( 1)

▼

Simplify right-hand side

CalcPow

Calculate power

h=-11±sqrt(121-4(1)(96))/2(1)

MultByOne

a * 1=a

h=-11±sqrt(121-4(96))/2

MultNegPos

(- a)b = - ab

h=-11±sqrt(121-384)/2

SubTerm

Subtract term

h=-11±sqrt(-263)/2

The discriminant of this quadratic expression is negative, so we do not have any more radical solutions to our equation. Now, since we know that the height equals 9, we can find the length and width. l=9-1=8 w=9+3=12 Therefore, the given prism has dimensions of 8* 12* 9.

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Exercises p.380-381

Exercise 16 Page 381

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