Class X: Polynomials

CLASS X: POLYNOMIALS SEBASTIAN VATTAMATTAM 1. Introduction 1.1. Polynomial. Example 1.1. Linear polynomial (1) p(x) = 3x − 8 (2) q(x) = −4x + 5 (3) f(x) = ax + b, a 6= 0 Example 1.2. Quadratic polynomial (1) p(x) = 3x2 − 8x + 2 (2) q(x) = −4x2 + 5x − 7 (3) f(x) = ax2 + bx + c, a 6= 0 Example 1.3. Cubic polynomial (1) p(x) = 3x3 − 8x2 + 2x − 1 (2) q(x) = −4x3 + 5x2 − 7x + 4 (3) f(x) = ax3 + bx2 + cx + d, a 6= 0 1.2. Zeros of a polynomial. Definition 1.4. If p(x) is a polynomial and p(k) = 0, then k is called a zero of the polynomial. Example 1.5. (1) p(x) = ax + b, a 6= 0 (2) p(−b a ) = 0 (3) −b a is a zero of p(x) (1) p(x) = x2 − 3x − 4 12 SEBASTIAN VATTAMATTAM (2) p(−1) = 0, p(4) = 0 (3) −1, 4 are zeros of p(x) Geometrical meaning y = p(x) represents a curve in the xy−plane. Zeros are the x-coordinates of the points at which the curve cuts the x-axis. (1) y = ax + b, a 6= 0 represents a line. It cuts the x-axis at (−b a , 0) −b a is a zero of p(x) = ax + b (2) y = x2 − 3x − 4 represents a parabola. It cuts the x-axis at (−1, 0), (4, 0) −1, 4 are zeros of p(x) = x2 −3x−4. If the parabola is totally above or below the x-axis, there will be no zero. If it only touches the x-axis, there will be only one zero. Theorem 1.6. If , are the zeros of p(x) = ax2 +bx+c, then + = −b a , .= ca 2. Division Algorithm Theorem 2.1. If p(x), g(x) are two polynomials, g(x) 6= 0, then there exist polynomials q(x) and r(x) such that (2.1) p(x) = g(x)q(x) + r(x), deg r(x) < deg g(x) Example 2.2. Divide 2x2 + 3x + 1 by x + 2CLASS X 3 2x -1 x +2 2x2 +3x +1 2x2 +4x -x +1 -x -2 3 Quotient = 2x − 1 Remainder = 3 Example 2.3. Divide 3x3 + x2 + 2x + 5 by 1 + 2x + x2 3x -5 x2 +2x +1 3x3 +x2 +2x +5 3x3 +6x2 +3x −5x2 -x +5 −5x2 -10x -5 9x +10 Quotient = 3x − 5. Remainder = 9x + 10 Definition 2.4. If p(x) = f(x).g(x) then f(x) and g(x) are factors of p(x) and p(x) is a multiple of each of them. Theorem 2.5. Remainder Theorem The remainder on dividing p(x) by x − a is p(a). Theorem 2.6. Factor theorem If p(x) is a polynomial and p(a) = 0, then x − a is a factor of p(x). By division algorithm, there exist q(x) and r(x) such that p(x) = (x − a)q(x) + r(x), deg r(x) < deg (x − a) Since deg r(x) < deg (x − a) = 1, r(x) is a constant polynommial Then p(a) = r(a) ) r(x) = 0 Therefore, p(x) = (x − a)q(x) ) (x − a) divides p(x). Problem 2.7. If p2,−p2 are zeros of 2x4 − 3x3 − 3x2 + 6x − 2, find all its zeros.4 SEBASTIAN VATTAMATTAM Let p(x) = 2x4 − 3x3 − 3x2 + 6x − 2 By factor theorem x − p2, x + p2 are factors of p(x) Therefore, (x − p2)(x + p2) = x2 − 2 is a factor. Let us divide p(x) by x2 − 2 2x2 -3x +1 x2 -2 2x4 −3x3 −x2 6x -5 2x4 −4x2 −3x3 x2 +6x -2 −3x3 +6x x2 -2 x2 -2 0 Thus p(x) = (x2 − 2)(2x2 − 3x + 1) 2x2 − 3x + 1 = (2x − 1)(x − 1) The zeros are p2,−p2, 1/2, 1 3. Synthetic Division Example 3.1. Divide 2x4 − x3 − 4x2 + x − 5 by x − 2 2 2 -1 -4 1 -5 0 4 6 4 10 2 3 2 5 5 Quotient = 2x3 + 3x2 + 2x + 5, Remainder = 5 Problem 3.2. x+6 is a factor of x3 +3x2 +4x+p. Find p.Let p(x) = x3 + 3x2 + 4x + p Method 1 Since x+6 is a factor of p(x), p(−6) = 0 ) (−6)3+3(−6)2+ 4(−6) + p = 0 ) p = 132 Method 2 -6 1 3 4 p 0 -6 18 -132 1 -3 22 p-132 p − 132 = 0 ) p = 132 E-mail address: vattamattam@gmail.com