Chapter 2: Polynomials

A highway underpass is parabolic in shape. The curve of the underpass is represented by the quadratic polynomial $f(x) = x^2 - 2x - 8$.

1. Find the zeroes of the polynomial $f(x)$.
   (A) 2, -4

   (B) 4, -2

   (C) -4, -2

   (D) 4, 2
   Show Solution
   Solution: (B) 4, -2
   Step 1: Set $x^2 - 2x - 8 = 0$.
   Step 2: Factorize: $x^2 - 4x + 2x - 8 = 0 \Rightarrow x(x-4) + 2(x-4) = 0$.
   Step 3: $(x-4)(x+2) = 0 \Rightarrow x = 4, x = -2$.
2. What is the distance between the two points where the arch meets the ground (x-axis)?
   (A) 4 units

   (B) 6 units

   (C) 8 units

   (D) 2 units
   Show Solution
   Solution: (B) 6 units
   Step 1: The points on the x-axis correspond to the zeroes: $4$ and $-2$.
   Step 2: Distance $= |4 - (-2)| = |4 + 2| = 6$ units.

A packaging company manufactures rectangular boxes. The volume of a box is given by the polynomial $V(x) = x^3 - 4x^2 + 3x$.

1. What are the possible dimensions (length, breadth, height) of the box in terms of $x$?
   (A) $x, x-1, x-3$

   (B) $x, x-2, x-3$

   (C) $x, x+1, x+3$

   (D) $x, x-1, x+3$
   Show Solution
   Solution: (A) $x, x-1, x-3$
   Step 1: Factorize $V(x) = x(x^2 - 4x + 3)$.
   Step 2: Factorize the quadratic part: $x^2 - 3x - x + 3 = x(x-3) - 1(x-3) = (x-1)(x-3)$.
   Step 3: Dimensions are $x, (x-1), (x-3)$.
2. If $x = 5$ units, what is the volume of the box?
   (A) 20 cubic units

   (B) 30 cubic units

   (C) 40 cubic units

   (D) 50 cubic units
   Show Solution
   Solution: (C) 40 cubic units
   Step 1: Substitute $x=5$ in $V(x) = x(x-1)(x-3)$.
   Step 2: $V(5) = 5(5-1)(5-3) = 5(4)(2) = 40$.

During a sports day event, a student throws a javelin. The path of the javelin is parabolic and is modeled by the polynomial $h(t) = -2t^2 + 8t + 10$, where $h$ is the height in meters and $t$ is the time in seconds.

1. What is the height of the javelin at $t = 1$ second?
   (A) 16 m

   (B) 18 m

   (C) 14 m

   (D) 12 m
   Show Solution
   Solution: (A) 16 m
   Step 1: Substitute $t=1$ in $h(t)$.
   Step 2: $h(1) = -2(1)^2 + 8(1) + 10 = -2 + 8 + 10 = 16$ meters.
2. At what time does the javelin hit the ground?
   (A) 1 s

   (B) 3 s

   (C) 5 s

   (D) 4 s
   Show Solution
   Solution: (C) 5 s
   Step 1: The javelin hits the ground when height $h(t) = 0$.
   Step 2: $-2t^2 + 8t + 10 = 0 \Rightarrow t^2 - 4t - 5 = 0$ (Dividing by -2).
   Step 3: Factorize: $(t-5)(t+1) = 0$.
   Step 4: $t = 5$ or $t = -1$. Since time cannot be negative, $t = 5$ seconds.
3. What is the maximum height achieved by the javelin?
   (A) 18 m

   (B) 20 m

   (C) 16 m

   (D) 22 m
   Show Solution
   Solution: (A) 18 m
   Step 1: The maximum height of a parabola $at^2 + bt + c$ occurs at $t = -b/2a$.
   Step 2: Here $a=-2, b=8$. So, $t = -8 / (2 \times -2) = -8 / -4 = 2$ seconds.
   Step 3: Find height at $t=2$: $h(2) = -2(2)^2 + 8(2) + 10 = -8 + 16 + 10 = 18$ meters.