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Polynomial -1 Class - 9

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Published in: Mathematics
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Kundan / Ranchi

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  1. 1. 2. 3. Polynomials (Part -1) Class IX Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x2 -3x+7 (ii) Y2 + 42 (iii) 34t + 612 (iv) y + 2/y (V) XIO + + t5() Answer (i) 4x2 - 3x + 7 There is only one variable x with whole number power so this polynomial in one variable. (ii) y 2 + 42 There is only one variable y with whole number power so this polynomial in one variable. (iii) 342 + There is only one variable t but in 34t power of t is 1/2 which is not a whole number so 34t + tN2 is not a polynomial. There is only one variable y but 2/ y = 2y-1 so the power is not a whole number so y + 2/ y is not a polynomial. (v) x10 + y3 + t50 There are three variable x, y and t and there powers are whole number so this polynomial in three variable. Write the coefficients of x2 in each of the following: (ii) 2 - x2 + x3 (iv) A12x - 1 Answer (i) coefficients of x2 = 1 (ii) coefficients of x (iii) coefficients of x2 = 7t/2 (iv) coefficients of x2 = 0 Give one example each of a binomial of degree 35, and of a monomial of degree 100. Answer 3x35+7 and 4x 100
  2. 4. 5. 6. Write the degree of each of the following polynomials: (i) 5x3 + 4x2 + 7x (ii) 4 —Y2 (iii) 5t — 47 (iv) 3 Answer (i) 5x3 has highest power in the given polynomial which power is 3. Therefore, degree of polynomial is 3. (ii) Y2 has highest power in the given polynomial which power is 2. Therefore, degree of polynomial (iii) 5t has highest power in the given polynomial which power is 1. Therefore, degree of polynomial is 1. (iv) There is no variable in the given polynomial. Therefore, degree of polynomial is 0. Classify the following as linear, quadratic and cubic polynomial: (i) x2 + x Quadratic Polynomial (ii) x - x3 Cubic Polynomial (iii) y + +4 Quadratic Polynomial (iv) 1 + x Linear Polynomial (v) 3t Linear Polynomial (vi) r2 Quadratic Polynomial (vii) 7x3 Cubic Polynomial Find the value of the polynomial at 5x + 4x2 + 3 at (i) x = 0 (ii) x = - 1 (iii) x = 2 Answer (i) p(x) = 5x + 4x2 + 3 = 5(0) + + 3
  3. (ii) p(x) = 5x + 4x2 + 3 p(-l) = 5(-1) + + 3 (iii) p(x) = 5x + 4x2 + 3 p(2) = 5(2) + + 3 = 10-16+3=-3 Find p(0), p(l) and p(2) for each of the following polynomials: 7. (i) p(y) = Y2 -y + 1 (ii) p(t) = 2 + t + 2t2 - t3 (iii) p(x) = x3 (iv)p(x) = (x - 1) (x + 1) Answer (i) p(y) = Y2 p(0) = (0)2 p(2) = (2)2 (ii) p(t) = 2 + t + 2t2 - t3 p(0) = 2+0+2 (0)2 - (0)3 = 2 - (1)3 p(2) = 2 + 2 + - (2)3 (iii) p(x) = x3 p(0) = (0)3 = O p(2) = (2)3 = 8 (iv) p(x) = 1(3) = 3 8. Verify whether the following are zeroes of the polynomial, indicated against them. (i)p(x) = 3x + 1, x = -1/3 (ii) p(x) = 5x - n, x = 4/5 (iii)p(x) = x2 - 1, x = 1, 1 (iv) p(x) = (x + 1) (x - (v) p(x) = x2 , x = 0 (vi) p(x) = Ix + m,x = 1 (vii) p(x) = 3x2 1, x = & + 1, x = 1/2 (viii) p(x) =
  4. Answer (i) If x = -1/3 is a zero of polynomial p(x) = 3x + 1 then p(-1/3) should be 0. At, p(-1/3) = 3(-1/3) + 1 - —-1+1=0 Therefore, x = -1/3 is a zero of polynomial p(x) = 3x + 1. (ii) If x = 4/5 is a zero of polynomial p(x) = 5x - 7t then p(4/5) should be 0. At, p(4/5) = 5(4/5) - = 4 - Therefore, x = 4/5 is not a zero of given polynomial p(x) = 5x - a. (iii) If x = 1 and x = -1 are zeroes of polynomial p(x) = x2 - 1, then p(l) and p(-l) should be 0. At, p(l) = (1)2 - 1 = 0 and 2-1=0 At, p(-l) = (-1) , — 1 and -1 are zeroes of the polynomial p(x) = x2 - 1. Hence x — (iv) If x = -1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x - 2), then p( - 1) and (2)should be 0. At p( 1) = , - (-1 + 1) (-1-2) = o = o, and At, p(2) = Therefore, x = -1 and x = 2 are zeroes of the polynomial p(x) = (x +1) (x - 2). (v) If x = 0 is a zero of polynomial p(x) = x2, then p(0) should be zero. Here, p(0) = (0)2 = O Hence, x = 0 is a zero of the polynomial p(x) = x2. = 3-— is a zero of polynomial p(x) = Ix + m then p (vi) If x should be 0. , = + is a zero of the polynomial (x) = Ix + m Therefore x —1 2 (vii) If x = are zeroes of polynomial p(x) = 3x2 + 1, then should be 0. -1 = 1 = 0 and = 3X2 1 , = — is a zero of polynomial p(x) Therefore x = is not a zero of this polyomial. but x (viii) If x = 1/2 is a zero of polynomial p(x) = 2x + 1 then p(1/2) should be 0.
  5. 9. At, p(1/2) = 2(1/2) + 1 Therefore, x = 1/2 is not a zero of given polynomial p(x) = 2x + 1. Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 (ii) p(x) = x 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x 2 (v) p(x) = 3x (vi) p(x) = (vii) p(x) = cx + d, c 0, c, are real numbers. Answer (i) p(x) = x + 5 p(x) = 0 Therefore, x = -5 is a zero of polynomial p(x) (ii) p(x) = x - 5 p(x) = 0 x-5=o Therefore, x = 5 is a zero of polynomial p(x) = x - 5. (iii) p(x) = 2x + 5 p(x) = 0 2x+5=o x = -5/2 Therefore, x = -5/2 is a zero of polynomial p(x) (iv) p(x) = 3x - 2 p(x) = 0 x = 2/3 Therefore, x = 2/3 is a zero of polynomial p(x) (v) p(x) = 3x p(x) = 0 Therefore, x = 0 is a zero of polynomial p(x) (vi) p(x) = ax p(x) = 0 ax = 0 =2x+5.
  6. Therefore, x = 0 is a zero of polynomial p(x) = (vii) p(x) = cx + d p(x) = 0 cx + d = O x = -d/c ax. Therefore, x = -d/c is a zero of polynomial p(x) = cx + d. Find the remainder when x3 + 3x2 + 3x + 1 is divided by 10. (ii) x - 1/2 (iii) x (iv) x + 7t (v) 5 + 2x Answer By long division, x3 3x2 3x I 2 2x2 + 3x I 2x2 + 2x Therefore, the remainder is 0. (ii) x - 1/2 By long division,
  7. 19 x3 3x2 3x I 19 19 19 27 Therefore, the remainder is 27/8. (iii) x x 2 + 3x+3 x3 + 3x2 + 3x + 1 3x2 + 3x + 1 3X2 Therefore, the remainder is 1. (iv) x + 7t
  8. x 2 + (3 — TT)x + (3 x3 3x2 3x I 2 2 — 37T + IT (3 — Tt)x2 + 3X + 1 (3 — TT)x2 + (3 — Tt)Ttx [3 — 3Tt + Tt2]X + 1 [3 — 3R + Tt2]x + (3 — 3TT + [1 — 3Tt + 3TT2 Therefore, the remainder is [1 - 37t + 37t x3 + 3x2 + 3x + 1 5 2 2 x 2 2 4 4 35 7 4 8 27 8 Therefore, the remainder is -27/8. Find the remainder when x3 - ax2 + 6x - a is divided by x - a. 11. Answer
  9. By Long Division, 6x — 6a x 182 + 6 ax 2 + 6x 6x a. Therefore, remainder obtained is 5a when x Check whether 7 + 3x is a factor of 3x3 + 7x. 12. Answer -ax + 6x- a is divided by x - a We have to divide 3x3 + 7x by 7 + 3x. If remainder comes out to be 0 then 7 + 3x will be a factor of 3x3 + By Long Division, 3x 7 70 3x3 + ox2 + 7x 3x3+ 7x2 -7x2 + 7x 49x 70x 70x 490 —490 As remainder is not zero so 7 + 3x is not a factor of 3x3 + 7x.

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