Solutions — Polynomials Topic Review

1. Fluency

   • (a) Degree and leading coefficient: Degree 4, leading coefficient 2.
   • (b) P(1) and P(−1): P(1) = 2−1+3−5 = −1. P(−1) = 2+1−3−5 = −5.
   • (c) Is (0, −5) on the graph?: P(0) = −5. Since P(0) = −5, the point (0,−5) does lie on the graph. ✓

2. Fluency

   • (a) Subtract polynomials: (3x³−2x²+x) − (x³+x²−4x+2) = 2x³−3x²+5x−2. 2x³ − 3x² + 5x − 2.
   • (b) Expand: x(x²+x−3) − 2(x²+x−3) = x³+x²−3x−2x²−2x+6 = x³ − x² − 5x + 6.

3. Fluency

   • (a) P(x)=x³+2x²−x+4, divisor (x−2): P(2) = 8+8−2+4 = 18.
   • (b) P(x)=3x³−x+5, divisor (x+1): P(−1) = −3+1+5 = 3.

4. Fluency

   • P(2) = 8−12+4 = 0. Since P(2) = 0, (x−2) is a factor of P(x) by the factor theorem. ✓

5. Fluency

   • x³÷x=x²; x²(x+3)=x³+3x²; subtract: x²−3x. Then x²÷x=x; x(x+3)=x²+3x; subtract: −6x−18. Then −6x÷x=−6; −6(x+3)=−6x−18; subtract: 0. x³+4x²−3x−18 = (x+3)(x²+x−6). Remainder = 0.

6. Understanding

   • (a) x³+x²−x−1: Test x=1: 1+1−1−1=0 ✓. Divide by (x−1): x²+2x+1=(x+1)². (x−1)(x+1)².
   • (b) x³−7x+6: Test x=1: 1−7+6=0 ✓. Divide by (x−1): x²+x−6=(x+3)(x−2). (x−1)(x+3)(x−2).

7. Understanding

   • (a) x-intercepts: x=−1, multiplicity 1 (crosses). x=2, multiplicity 2 (touches and bounces).
   • (b) y-intercept: P(0)=−(1)(−2)²=−(1)(4)=−4. y-intercept: (0, −4).
   • (c) End behaviour: Degree 3, negative leading coefficient. As x→−∞: y→+∞. As x→+∞: y→−∞. (Rises left, falls right.)
   • (d) Behaviour at x=2: x=2 is a double zero, so the graph touches the x-axis and turns back without crossing.

8. Understanding

   • (a) Equation: P(2) = 6: 8+4k−6+2 = 6 ⇒ 4k+4 = 6 ⇒ 4k = 2.
   • (b) Solve for k: k = 2÷4 = k = ½.

9. Understanding

   • (a) Test x=−1 and x=2: P(−1)=1+2−3−4+4=0 ✓. P(2)=16−16−12+8+4=0 ✓. Both are zeros.
   • (b) Factored form: Both are double zeros ⇒ P(x)=(x+1)²(x−2)². Verify: [(x+1)(x−2)]²=(x²−x−2)²=x⁴−2x³−3x²+4x+4 ✓
   • (c) Behaviour and end: At x=−1 and x=2: graph touches and bounces (double zeros). Degree 4, positive leading coefficient: both ends rise (y→+∞ as x→±∞).

10. Problem Solving

    • (a) Two equations: P(1)=0: 1+a−b−6=0 ⇒ a−b=5. P(−2)=0: −8+4a+2b−6=0 ⇒ 4a+2b=14 ⇒ 2a+b=7.
    • (b) Solve: Add the equations: 3a=12 ⇒ a=4. Then b=2(4)+7−4(1)... using a−b=5: b=4−5=−1. a=4, b=−1.
    • (c) Factored form: P(x)=x³+4x²−x−6. (x−1) and (x+2) are factors, so (x−1)(x+2)=x²+x−2. Divide: x³+4x²−x−6 ÷ (x²+x−2) = x+3. P(x)=(x−1)(x+2)(x+3).

11. Problem Solving

    • (a) General form: Zeros at x=−3 (simple) and x=1 (double): P(x) = a(x+3)(x−1)².
    • (b) Find a: P(2)=5: a(5)(1)=5 ⇒ a = 1.
    • (c) Complete polynomial: P(x) = (x+3)(x−1)².
    • (d) y-intercept: P(0) = (3)(1) = 3. y-intercept: (0, 3).

12. Problem Solving

    • (a) Two equations: P(−1)=0: −1+a−b+6=0 ⇒ a−b=−5. P(5)=36: 125+25a+5b+6=36 ⇒ 25a+5b=−95 ⇒ 5a+b=−19.
    • (b) Solve: Add both equations: 6a=−24 ⇒ a=−4. From a−b=−5: b=a+5=1. a=−4, b=1.
    • (c) Factorise: P(x)=x³−4x²+x+6. Divide by (x+1): quotient x²−5x+6=(x−2)(x−3). P(x)=(x+1)(x−2)(x−3).
    • (d) Sketch: x-intercepts: −1, 2, 3 (all simple, all cross). y-intercept: P(0)=6. Degree 3, positive leading coefficient: falls left, rises right. Curve crosses at x=−1, rises to local max, dips to local min between x=2 and x=3, then rises.