Solutions — Introduction to Polynomials

1. Degree and leading coefficient. Fluency

   • (a) 3x²−2x+5: Degree 2, leading coefficient 3. (Quadratic.)
   • (b) x⁵−4x³+2: Degree 5, leading coefficient 1. (Degree-5 polynomial.)
   • (c) −2x³+x−7: Degree 3, leading coefficient −2. (Cubic.)
   • (d) 6x⁴−3x²+x: Degree 4, leading coefficient 6. (Quartic.)

2. Standard form. Fluency

   • (a) 3+2x−x²: −x²+2x+3, degree 2.
   • (b) 4x³−2+x−5x²: 4x³−5x²+x−2, degree 3.
   • (c) x+3x⁴−x²: 3x⁴−x²+x, degree 4.
   • (d) 5−3x+7x²−x³: −x³+7x²−3x+5, degree 3.

3. Evaluate the polynomial. Fluency

   • (a) P(0): P(0)=0−0+0−4=−4.
   • (b) P(1): P(1)=2−3+1−4=−4.
   • (c) P(−1): P(−1)=2(−1)−3(1)+(−1)−4=−2−3−1−4=−10.
   • (d) P(2): P(2)=2(8)−3(4)+2−4=16−12+2−4=2.

4. Add and subtract polynomials. Fluency

   • (a) (3x²−2x+1)+(x²+4x−3): Collect like terms: (3+1)x²+(−2+4)x+(1−3) = 4x²+2x−2.
   • (b) (5x³−2x+4)−(2x³+x²−3x+1): Distribute the minus: 5x³−2x+4−2x³−x²+3x−1 = 3x³−x²+x+3.
   • (c) (2x⁴−x²+3x)+(x³+4x²−x−5): 2x⁴+x³+3x²+2x−5.
   • (d) (x³−4x²+2x−1)−(x³+2x−5): x³−4x²+2x−1−x³−2x+5 = −4x²+4.

5. Multiply polynomials. Understanding

   • (a) x(x²−3x+2): x³−3x²+2x.
   • (b) (x+2)(x²−x+3): x(x²−x+3)+2(x²−x+3) = x³−x²+3x+2x²−2x+6 = x³+x²+x+6.
   • (c) (x−1)(x+1)(x−2): First: (x−1)(x+1)=x²−1. Then (x²−1)(x−2)=x³−2x²−x+2. x³−2x²−x+2.
   • (d) (2x+1)(x²−2x+3): 2x(x²−2x+3)+1(x²−2x+3)=2x³−4x²+6x+x²−2x+3 = 2x³−3x²+4x+3.

6. End behaviour. Understanding

   • (a) y=x³−2x+1: Degree 3 (odd), leading coeff +1 (positive). As x→−∞: y→−∞. As x→+∞: y→+∞.
   • (b) y=−x⁴+3x²: Degree 4 (even), leading coeff −1 (negative). As x→±∞: y→−∞ (both ends go down).
   • (c) y=2x⁵−x³: Degree 5 (odd), leading coeff +2 (positive). As x→−∞: y→−∞. As x→+∞: y→+∞.
   • (d) y=−3x²+5x−1: Degree 2 (even), leading coeff −3 (negative). As x→±∞: y→−∞ (opens downward).

7. Polynomial from factored form. Understanding

   • (a) x-intercepts: Set each factor to zero: x = 1, x = −2, x = 3.
   • (b) Standard form: (x−1)(x+2)=x²+x−2. Then (x²+x−2)(x−3)=x³−3x²+x²−3x−2x+6 = x³−2x²−5x+6.
   • (c) y-intercept: P(0)=(−1)(2)(−3)=6. y-intercept: (0, 6).
   • (d) End behaviour: Degree 3 (odd), leading coeff +1. As x→−∞: y→−∞. As x→+∞: y→+∞. Falls left, rises right.

8. Does the point lie on the polynomial? Understanding

   • (a) (0, 2): P(0)=0−0−0+2=2 ✓. Yes.
   • (b) (1, 0): P(1)=1−2−1+2=0 ✓. Yes.
   • (c) (2, 1): P(2)=8−8−2+2=0 ≠ 1. No. (The point (2, 0) lies on the graph, not (2, 1).)
   • (d) (−1, 0): P(−1)=−1−2+1+2=0 ✓. Yes.

9. Find the unknown coefficient. Problem Solving

   • P(2)=k(8)−3(4)+2(2)−1=8k−12+4−1=8k−9. Set equal to 5: 8k−9=5 ⇒ 8k=14 ⇒ k = 7/4.

10. Construct a polynomial from zeros. Problem Solving

    • (a) Factored form: x-intercepts at −2, 0, 3 ⇒ factors (x+2), x, (x−3). P(x) = k ⋅ x(x+2)(x−3)
    • (b) Find k: At (1,12): 12 = k(1)(3)(−2) = −6k ⇒ k = −2.
    • (c) Standard form: P(x)=−2x(x+2)(x−3)=−2x(x²−x−6)=−2x³+2x²+12x.