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Topics · Polynomials & Remainder Theorem

Polynomials in DSE Maths Paper 2

Polynomials are the algebra the rest of the paper is built on. Questions ask you to expand and simplify products, factorise expressions, use the remainder and factor theorems to find unknown coefficients, and work with the H.C.F. and L.C.M. of algebraic expressions.

What this topic covers

  • Expanding and simplifying products of polynomials
  • Factorising, including by grouping and by the factor theorem
  • The remainder theorem: finding a remainder or an unknown coefficient
  • H.C.F. and L.C.M. of polynomials in factorised form

Two sample questions

Sample 1

(7x squared plus 10x plus 5)(x minus 6) equals

  1. A7x cubed plus 10x squared plus 5x minus 30.
  2. B7x cubed minus 32x squared minus 55x minus 30.
  3. C7x cubed plus 5x squared plus 10x minus 30.
  4. D7x cubed plus 52x squared minus 55x minus 30.
Show the answer and solution

Answer: B

Worked solution

(7x2+10x+5)(x−6)\displaystyle (7x^{2} + 10x + 5)(x - 6)=7x3\displaystyle {}= 7x^{3}−42x2\displaystyle {}- 42x^{2}+10x2\displaystyle {}+ 10x^{2}−60x\displaystyle {}- 60x+5x\displaystyle {}+ 5x−30\displaystyle {}- 30=7x3−32x2−55x−30\displaystyle = 7x^{3} - 32x^{2} - 55x - 30

Sample 2

The H.C.F. and the L.C.M. of the three polynomials f(x), g(x) and h(x) are 3(x minus 5)(x plus 4) squared and 12(x minus 5)(x minus 1) squared (x plus 4) cubed respectively. If f(x) equals 3(x minus 5)(x minus 1)(x plus 4) squared and g(x) equals 6(x minus 5)(x minus 1) squared (x plus 4) squared, then h(x) is

  1. A3(x minus 5)(x plus 4) cubed .
  2. B3(x minus 5)(x minus 1) squared (x plus 4) squared .
  3. C12(x plus 4) cubed .
  4. D12(x minus 5)(x plus 4) cubed .
Show the answer and solution

Answer: D

Worked solution

The H.C.F. takes the smallest power of each factor. Its power of (x−1)\displaystyle (x - 1) is lower than in both f(x)\displaystyle f(x) and g(x)\displaystyle g(x), so h(x)\displaystyle h(x) has no factor (x−1)\displaystyle (x - 1).The L.C.M. takes the largest power of each factor. Its power of (x+4)\displaystyle (x + 4) is higher than in both f(x)\displaystyle f(x) and g(x)\displaystyle g(x), so h(x)\displaystyle h(x) has (x+4)3\displaystyle (x + 4)^{3}.f(x)\displaystyle f(x) and g(x)\displaystyle g(x) both have (x−5)\displaystyle (x - 5), the same power as the H.C.F. and the L.C.M., so h(x)\displaystyle h(x) has (x−5)\displaystyle (x - 5) too.In the same way, the H.C.F. of 3,6\displaystyle 3, 6 and the coefficient k\displaystyle k of h(x)\displaystyle h(x) is 3\displaystyle 3 and their L.C.M. is 12=22×3\displaystyle 12 = 2^{2} \times 3, so k=22×3=12\displaystyle k = 2^{2} \times 3 = 12.∴h(x)=12(x−5)(x+4)3\displaystyle \therefore h(x) = 12(x - 5)(x + 4)^{3}

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Original material by Dsetopia, written from the EDB syllabus learning objectives. Not affiliated with the HKEAA; no exam content is reproduced.