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03.05 · Algebra

Polynomials & Cubic Equations

Elite lesson11 formulas10 questions

On the syllabus of

Elite lesson19 minEnglish

Formulas

At a glance 11 formulas

  1. 1Remainder theoremP(x)÷(x−a) leaves P(a)P(x) \div (x - a) \text{ leaves } P(a)P(x)÷(ax−b) leaves P(ba)P(x) \div (ax - b) \text{ leaves } P\left(\tfrac{b}{a}\right)
  2. 2Factor theorem(x−a) is a factor of P(x)  ⟺  P(a)=0(x - a) \text{ is a factor of } P(x) \iff P(a) = 0
  3. 3Division algorithmP(x)=D(x) Q(x)+R(x)P(x) = D(x)\,Q(x) + R(x)D divisor, Q quotient, R remainder, deg⁡R<deg⁡DD \text{ divisor},\ Q \text{ quotient},\ R \text{ remainder},\ \deg R < \deg D
  4. 4Remainder by a quadraticP(x)÷(x−a)(x−b) leaves px+qP(x) \div (x - a)(x - b) \text{ leaves } px + qpa+q=P(a), pb+q=P(b)pa + q = P(a),\ pb + q = P(b)
  5. 5Cubic: sum and productr1+r2+r3=−ba, r1r2r3=−dar_1 + r_2 + r_3 = -\dfrac{b}{a},\ r_1r_2r_3 = -\dfrac{d}{a}r1r2+r2r3+r3r1=ca; r1,r2,r3 the roots of ax3+bx2+cx+d=0r_1r_2 + r_2r_3 + r_3r_1 = \dfrac{c}{a};\ r_1, r_2, r_3 \text{ the roots of } ax^3 + bx^2 + cx + d = 0
  6. 6Sum of reciprocals1r1+1r2+1r3=−cd\dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} = -\dfrac{c}{d}r1,r2,r3 roots of ax3+bx2+cx+d=0r_1, r_2, r_3 \text{ roots of } ax^3 + bx^2 + cx + d = 0
  7. 7Factored formP(k)=a(k−r1)(k−r2)(k−r3)P(k) = a(k - r_1)(k - r_2)(k - r_3)r1,r2,r3 the roots of P, a its leading coefficientr_1, r_2, r_3 \text{ the roots of } P,\ a \text{ its leading coefficient}
  8. 8Roots in APIf a cubic's roots are in AP, write them m−dm - d, mm, m+dm + d: the sum of roots gives mm at once, and mm is itself a root.
  9. 9Values on a lineIf a cubic PP has P(1),P(2),P(3)P(1), P(2), P(3) on a line mx+cmx + c, then P(x)−(mx+c)P(x) - (mx + c) is 0 at 1, 2 and 3, so it is (x−1)(x−2)(x−3)(x - 1)(x - 2)(x - 3) times PP's leading coefficient.
  10. 10Surd inputTo evaluate P(x)P(x) at x=a+bx = a + \sqrt{b}, use (x−a)2=b(x - a)^2 = b to cut the powers down before substituting.
  11. 11Compare coefficientsWhen two polynomials are equal for every xx, match the coefficients of each power: x3x^3 with x3x^3, x2x^2 with x2x^2, down to the constants.

Test yourself

10 questions · 3 easy · 4 medium · 3 hard

Tap an option to answer. The first tap counts.

  1. Q1Easy

    Long division of P(x)P(x) by x−3x - 3 gives the quotient x2+2x−1x^2 + 2x - 1 and the remainder 5. Find P(2)P(2).

    Correct answer−2-2

    Why

    Write P(x)=(x−3)(x2+2x−1)+5P(x) = (x - 3)(x^2 + 2x - 1) + 5 and substitute directly.

    At x=2x = 2: (2−3)(4+4−1)+5=−7+5(2 - 3)(4 + 4 - 1) + 5 = -7 + 5.

    So P(2)=−2P(2) = -2.

  2. Q2Easy

    Asked in IPMAT Indore 2023

    A polynomial p(x)p(x) leaves a remainder 2 when divided by (x−1)(x - 1) and a remainder 1 when divided by (x−2)(x - 2). The remainder when p(x)p(x) is divided by (x−1)(x−2)(x - 1)(x - 2) is

    Correct answer3−x3 - x

    Why

    A quadratic divisor can only leave a remainder of degree 1 or less, say px+qpx + q.

    The two given remainders are its values: p+q=2p + q = 2 at x=1x = 1 and 2p+q=12p + q = 1 at x=2x = 2, so p=−1p = -1 and q=3q = 3.

    That makes the remainder 3−x3 - x.

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