University:
Texas A&M University
Course:
MATH 102 | Algebra
Academic year:

2020
Views:

345

Pages:

3
Author:

Alma Stephenson

MATHS POLYNOMIALS TYPES OF POLYNOMIALS TYPES OF POLYNOMIALS (A) BASED ON DEGREE: If degree of polynomial is Examples (B) 1 One Linear x + 3, y – x + 2, 3 x –3 2 Two Quadratic 2x2 –7, x2+y2 –2xy, x2 +1+ 3y 3 Three Cubic x3 + 3x2 –7x+8, 2x2+5x3+7, 4 Four bi-quadratic x4 + y4 + 2x2y2, x4 + 3,… BASED ON TERMS: 1 3 If number of terms in polynomial is Examples 7 x16, xy, …… 3 1. One Monomial 2. Two Binomial 2 + 7y6, y3 + x14, 7 + 5x9,… Three Trinomial x3 –2x + y, x31+y32+ z33,….. 3. 7x, 5x9, 1 MATHS Note : (1) Degree of constant polynomials (Ex.5, 7, –3, 8/5, …) is zero. (2) Degree of zero polynomial (zero = 0 = zero polynomial) is not defined. POLYNOMIAL IN ONE VARIABLE If a polynomial has only one variable then it is called polynomial in one variable. Ex. P(x) = 2x3 + 5x – 3 Cubic trinomial Q(x) = 7x7 – 5x5 – 3x3 + x + 3 polynomial of degree 7 R(y) = y Linear, monomial S(t) = t2 + 3 Quadratic Binomial Note: General form of a polynomial in one variable x of degree 'n' is anxn + an–1xn–1 + an–2xn–2 + ….+ a2x2 + a1x + a0, an ¹ 0, where an, an–1,… a2, a1, a0 all are constants. for linear ax + b, a¹0 for quadratic ax2 + bx + c, a¹0 for cubic ax3 + bx2 + cx + d, a¹0 REMAINDER THEOREM (i) Remainder obtained on dividing polynomial p(x) by x – a is equal to p(a) . (ii) If a polynomial p(x) is divided by (x + a) the remainder is the value of p(x) at x = –a. (iii) (x – a) is a factor of polynomial p(x) if p(a) = 0 (iv) (x + a) is a factor of polynomial p(x) if p(–a) = 0 (v) (x – a) (x – b) is a factor of polynomial p(x), if p(a) = 0 and p(b) = 0. 2 MATHS Ex.1 Find the remainder when 4x3 – 3x2 + 2x – 4 is divided by (a) x – 1 (b) x + 2 (c) x + 1 2 Sol. Let p(x) = 4x3 – 3x2 + 2x – 4 (a) When p(x) is divided by (x – 1), then by remainder theorem, the required remainder will be p(1) p(1) = 4 (1)3 – 3(1)2 + 2(1) – 4 =4×1–3×1+2×1–4 =4–3+2–4=–1 (b) When p(x) is divided by (x + 2), then by remainder theorem, the required remainder will be p (–2). p(–2) = 4 (–2)3 – 3 (–2)2 + 2(–2) – 4 = 4 × (–8) – 3 × 4 – 4 – 4 = – 32 – 12 – 8 = – 52 (c) 1 When p(x) is divided by, x +  then by remainder theorem, the required  2 remainder will be 1 1 1 1 p −  = 4 −  – 3 −  + 2 −  – 4 3  2  2 2  2  2 1 1 1 = 4 × −  – 3 × – 2 × – 4  8 1 2 =– – = 4 2 3 – 1– 4 = 1 – 3 – 5 4 2 4 −2−3−20 = −25 4 4 3