Solutions — Dividing Polynomials & the Remainder Theorem

1. Long division. Fluency

   • (a) (x²−5x+6)÷(x−2): x²−5x+6 = (x−2)(x−3) + 0. The divisor is a factor; remainder = 0.
   • (b) (x³−2x²+x−3)÷(x−1): Divide: x³÷x=x²; multiply x²(x−1)=x³−x²; subtract to get −x²+x. Divide: −x²÷x=−x; multiply −x(x−1)=−x²+x; subtract to get 0−3. Remainder = −3. x³−2x²+x−3 = (x−1)(x²−x) + (−3).
   • (c) (x³+x²−4x−4)÷(x+2): x³÷x=x²; x²(x+2)=x³+2x²; subtract: −x²−4x. Then −x²÷x=−x; −x(x+2)=−x²−2x; subtract: −2x−4. Then −2x÷x=−2; −2(x+2)=−2x−4; subtract: 0. x³+x²−4x−4 = (x+2)(x²−x−2) + 0.
   • (d) (2x³−x²+3x−1)÷(x−1): 2x³÷x=2x²; 2x²(x−1)=2x³−2x²; subtract: x²+3x. Then x²÷x=x; x(x−1)=x²−x; subtract: 4x−1. Then 4x÷x=4; 4(x−1)=4x−4; subtract: 3. Remainder = 3. 2x³−x²+3x−1 = (x−1)(2x²+x+4) + 3.

2. Remainder theorem. Fluency

   • (a) P(x)=x³−2x²+x−5, divisor (x−3): Remainder = P(3) = 27−18+3−5 = 7.
   • (b) P(x)=2x³+x²−x+4, divisor (x+1): Remainder = P(−1) = 2(−1)+1+1+4 = −2+1+1+4 = 4.
   • (c) P(x)=x⁴−3x²+2, divisor (x−2): Remainder = P(2) = 16−12+2 = 6.
   • (d) P(x)=x³−4x+1, divisor (x+2): Remainder = P(−2) = −8+8+1 = 1.

3. Factor theorem. Fluency

   • (a) P(x)=x³−x²−4x+4, test (x−2): P(2)=8−4−8+4=0. Yes, (x−2) is a factor.
   • (b) P(x)=x³+2x²−x−2, test (x+2): P(−2)=−8+8+2−2=0. Yes, (x+2) is a factor.
   • (c) P(x)=x³−3x²+x−3, test (x−3): P(3)=27−27+3−3=0. Yes, (x−3) is a factor.
   • (d) P(x)=x³−2x²+x−4, test (x−1): P(1)=1−2+1−4=−4≠0. No, (x−1) is not a factor.

4. Find the unknown coefficient. Fluency

   • (a) P(x)=x²+kx−6, factor (x−2): P(2)=0: 4+2k−6=0 ⇒ 2k=2 ⇒ k = 1.
   • (b) P(x)=x³−kx²+x−4, factor (x+1): P(−1)=0: −1−k(−1)+−1−4=0 ⇒ −1+k−1−4=0 ⇒ Wait — P(−1)=(−1)³−k(−1)²+(−1)−4=−1−k−1−4=−6−k=0 ⇒ k=−6. k = −6.
   • (c) P(x)=2x³+x²−kx+3, factor (x−3): P(3)=0: 2(27)+9−3k+3=0 ⇒ 54+9+3−3k=0 ⇒ 66=3k ⇒ k = 22.
   • (d) P(x)=x³+kx−2, factor (x−1): P(1)=0: 1+k−2=0 ⇒ k=1. k = 1.

5. Fully factorise. Understanding

   • (a) P(x)=x³−6x²+11x−6: Test x=1: P(1)=1−6+11−6=0 ✓. Divide by (x−1): quotient x²−5x+6=(x−2)(x−3). (x−1)(x−2)(x−3).
   • (b) P(x)=x³+2x²−5x−6: Test x=2: P(2)=8+8−10−6=0 ✓. Divide by (x−2): quotient x²+4x+3=(x+1)(x+3). (x−2)(x+1)(x+3).
   • (c) P(x)=x³−4x²+x+6: Test x=−1: P(−1)=−1−4−1+6=0 ✓. Divide by (x+1): quotient x²−5x+6=(x−2)(x−3). (x+1)(x−2)(x−3).
   • (d) P(x)=2x³−x²−2x+1: Test x=1: P(1)=2−1−2+1=0 ✓. Divide by (x−1): quotient 2x²+x−1=(2x−1)(x+1). (x−1)(2x−1)(x+1).

6. Build the polynomial from its zeros. Understanding

   • (a) Factored form: Zeros at x=−1, 2, 3 ⇒ P(x) = (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=x³−4x²+x+6. So a=−4, b=1, c=6.
   • (c) Verify y-intercept: P(0)=(1)(−2)(−3)=6. The constant term c=6. They match. ✓
   • (d) P(4): P(4)=(5)(2)(1)=10.

7. Find k using the remainder theorem. Understanding

   • (a) Equation: P(2)=12. So 8−12+2k+4=12 ⇒ 2k=12 ⇒ k=6. Equation: 2k = 12.
   • (b) Solve for k: 2k=12 ⇒ k = 6.
   • (c) Verify: P(x)=x³−3x²+6x+4. P(2)=8−12+12+4=12 ✓

8. Possible remainders. Understanding

   • (a) 5: Yes. Degree 0 < degree 2 of divisor. A constant remainder is always valid.
   • (b) x²−2: No. Degree 2 is equal to the degree of the divisor. The remainder must have degree strictly less than 2.
   • (c) 3x+1: Yes. Degree 1 < degree 2. A linear remainder is valid when dividing by a quadratic.
   • (d) x²+3x: No. Degree 2 equals the degree of the divisor. The remainder must have degree strictly less than 2.

9. Find a and b from two factor conditions. Problem Solving

   • (a) Two equations: P(3)=0: 9+3a+b=0 ⇒ 3a+b=−9. P(−2)=0: 4−2a+b=0 ⇒ −2a+b=−4. 3a+b=−9 and −2a+b=−4.
   • (b) Solve: Subtract: (3a+b)−(−2a+b)=−9−(−4) ⇒ 5a=−5 ⇒ a=−1. Then b=−4+2(−1)=−6. a = −1, b = −6.
   • (c) Factored form: x²−x−6 = (x−3)(x+2). Verify: (3−3)(3+2)=0 ✓; (−2−3)(−2+2)=0 ✓.

10. Two conditions — find a, b, and fully factorise. Problem Solving

    • (a) Two equations: Factor: P(2)=0: 8+4a+2b−12=0 ⇒ 4a+2b=4 ⇒ 2a+b=2. Remainder: P(1)=−6: 1+a+b−12=−6 ⇒ a+b=5.
    • (b) Solve: Subtract: (2a+b)−(a+b)=2−5 ⇒ a=−3. Then b=5−(−3)=8. a = −3, b = 8.
    • (c) Factorise: P(x)=x³−3x²+8x−12. Divide by (x−2): quotient x²−x+6. Discriminant: 1−24=−23<0, so x²−x+6 has no real factors. P(x) = (x−2)(x²−x+6) (fully factorised over the reals).