Question

PLEASE SHOW FULL SOLUTIONS FOR BOTH QUESTIONS AND ONLY ANSWER IF YOU KNOW! NO CALCULUS PLEASE! THAT WOULD BE VERY APPRECIATED!!

Answer 1

5)The dimensions of the rectangular prism are 3 , 7 and 2.

6) f(x) = x³ - 3x² - x + 3

Explanation:

5) The volume of rectangular prism = l*w*h.

l*w*h= 42

(x- 1)(x -2)(x+3) = 42

We can find (x - 2) *(x + 3) using the identity (x + a)(x + b) = x² + (a +b)*x + ab

(x - 2)(x + 3) =x² + (-2 +3)x + (-3)*2

= x² + 1x - 6

(x -1)(x + 3)(x - 2) = 42

(x - 1)(x² + x - 6) = 42

x*x² + x*x - 6*x + (-1)*x² + (-1) *x + (-1)(-6) = 42

x³ + x² - 6x - x² - x + 6 - 42 = 0

x³ + x² - x² - 6x - x + 6 -42 = 0

Combine the like terms,

x³ - 7x - 36 = 0

Find the zeros of the cubic polynomial by synthetic division method.

4 1 0 -7 -36

4 16 36

1 4 9 0

x - 4 is zero of the polynomial.

Ignore the quadratic polynomial x² + 4x + 9 as it will have irrational roots(zeros) and dimensions will be always positive integer.

x - 4 = 0

x = 4

length = x - 1 = 4 - 1 = 3

Width = x + 3 = 4 + 3 = 7

Height = x - 2 = 4 - 2 = 2

The dimensions of the rectangular prism are 3 , 7 and 2.

6) f(x) = x³ + bx² + cx + d

It is given that when f(x) is dived by x + 1 and x- 1, the remainders are same.

x + 1 = 0 ; x - 1 = 0

x = -1 ; x = 1

f(-1) = f(1)

-1 + b - c + d = 1 + b + c + d

-1 -1 + b - b - c - c + d - d = 0

-2 - 2c = 0

-2c = 2

c = 2 ÷ (-2)

`\boxed{c = -1}` -------------(I)

It is given that when f(x) divided by ( x - 2) it leaves a remainder (-3)

f(2) = -3

8 + 4b + 2c + d = -3

8 + 4b + 2*(-1) + d = -3 {from (I)}

8 + 4b - 2 + d = -3

4b + d = -3 + 2 - 8

4b + d = -9 -------------(II)

It is given that when f(x) divided by ( x + 3) it leaves a remainder (-48).

f(-3) = -48

-27 + 9b - 3c + d = -48

-27 + 9b - 3*(-1) +d = -48 {From (I)}

-27 + 9b + 3 + d = - 48

9b + d = -48 + 27 - 3

9b + d = -24 --------------(III)

Subtract equation (III) from equation (II),

(II) 4b + d = -9

(III) 9b + d = -24

- - +

-5b = 15

b = 15 ÷ (-5)

`\boxed{b=-3}`

Plugin b = -3 in equation (II),

4*(-3) + d = -9

-12 + d = -9

d = -9 + 12

`\boxed{d = 3}`

`\boxed{\bf f(x) = x^3 - 3x^2 -x + 3 }`

Answer 2

5. 3 cm × 2 cm × 7 cm

6. f(x) = x³ -3x² -x +3

Explanation:

You want the dimensions of a cuboid with edges marked (x-1), (x-2), and (x+3) and a volume of 42 cm³. You also want the coefficients of a monic cubic f(x) with values f(2) = -3, f(-3) = -48, and f(-1) = f(1).

5. Cuboid

The volume of the cuboid is the product of its dimensions, so we have ...

V = LWH

42 = (x -1)(x -2)(x +3)

The value of x can be found as the solution to this equation. A graphical solution is attached. It shows x=4, so the dimensions are ...

4 -1 = 3

4 -2 = 2

4 +3 = 7

The dimensions of the cuboid are 3 cm by 2 cm by 7 cm.

Note that the prime factors of 42 are 2, 3, 7, which have the required differences. You don't really need a polynomial to guess these are the dimensions.

The expanded polynomial is x³ -7x -36 = 0, so potential rational roots will be from the set {1, 2, 3, 4, 6, 9, 12, 18, 36}. An estimate of the upper bound of the real root puts it at ∛36 +√7 ≈ 5.9. We require x > 2, so the viable choices for testing are 3 and 4. x=4 is the solution.

6. Coefficients

The remainder theorem tells you that the remainder from dividing f(x) by (x-a) is f(a). To obtain linear equations in b, c, d, we can rearrange the function to ...

bx² +cx +d = -x³ +f(x)

For x = ±1, we know the remainders f(x) are the same, so we can look at the difference of the equations for these x-values.

(b(-1)² +c(-1) +d) - (b(1)² +c(1) +d) = (-(-1)³ +f(-1)) - (-(1)³ +f(1))

b -b -c -c +d -d = 1 -(-1)

-2c = 2

We can fill in values x=2, x=-3 to get two more equations in b, c, d. The coefficients of the three equations we have for the three unknowns are shown in the second attachment. The third attachment shows the solution of these equations is (b, c, d) = (-3, -1, 3).

The table in the first attachment confirms the remainders using these coefficients.

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Additional comment

For problem 6, we started out using synthetic division, then realized the resulting equations are the same as those developed using the rearranged form shown above. We like to let calculators and spreadsheets do the tedious arithmetic where possible.