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#mathsclass9 #rajansir Exercise 2.3 Class 9 Maths | Factor Theorem Exercise 2.3 Class 9 Maths | Factor Theorem by @rajansir07 online class 9811757843 Class 9 Maths Exercise 2.3 Chapter 2 Polynomials Class 9 Class 9 Maths NCERT Solutions Exercise 2.3 Polynomials Exercise 2.3 Class 9 Class 9 Maths Chapter 2 Exercise 2.3 Solutions Polynomial Factorization Class 9 Class 9 Maths Full Chapter 2 Solutions CBSE Class 9 Maths Polynomials How to solve Exercise 2.3 Class 9 Easy way to solve Polynomials Class 9 #Class9Maths #Exercise2_3 #Polynomials #NCERTSolutions #MathsShorts #CBSEClass9 #MathsTricks #BoardExamTips Class 9 Chapter 2 – Polynomials Exercise 2.4 1. Determine which of the following polynomials has (x + 1) a factor: (i) x3+x2+x+1 Solution: Let p(x) = x3+x2+x+1 The zero of x+1 is -1. [x+1 = 0 means x = -1] p(−1) = (−1)3+(−1)2+(−1)+1 = −1+1−1+1 = 0 ∴By factor theorem, x+1 is a factor of x3+x2+x+1 (ii) x4+x3+x2+x+1 Solution: Let p(x)= x4+x3+x2+x+1 The zero of x+1 is -1. . [x+1= 0 means x = -1] p(−1) = (−1)4+(−1)3+(−1)2+(−1)+1 = 1−1+1−1+1 = 1 ≠ 0 ∴By factor theorem, x+1 is not a factor of x4 + x3 + x2 + x + 1 (iii) x4+3x3+3x2+x+1 Solution: Let p(x)= x4+3x3+3x2+x+1 The zero of x+1 is -1. p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1 =1−3+3−1+1 =1 ≠ 0 ∴By factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1 (iv) x3 – x2– (2+√2)x +√2 Solution: Let p(x) = x3–x2–(2+√2)x +√2 The zero of x+1 is -1. p(−1) = (-1)3–(-1)2–(2+√2)(-1) + √2 = −1−1+2+√2+√2 = 2√2 ≠ 0 ∴By factor theorem, x+1 is not a factor of x3–x2–(2+√2)x +√2 2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x). (ii) p(x)=x3+3x2+3x+1, g(x) = x+2 Solution: p(x) = x3+3x2+3x+1, g(x) = x+2 g(x) = 0 ⇒ x+2 = 0 ⇒ x = −2 ∴ Zero of g(x) is -2. Now, p(−2) = (−2)3+3(−2)2+3(−2)+1 = −8+12−6+1 = −1 ≠ 0 ∴By factor theorem, g(x) is not a factor of p(x). (iii) p(x)=x3–4x2+x+6, g(x) = x–3 Solution: p(x) = x3–4x2+x+6, g(x) = x -3 g(x) = 0 ⇒ x−3 = 0 ⇒ x = 3 ∴ Zero of g(x) is 3. Now, p(3) = (3)3−4(3)2+(3)+6 = 27−36+3+6 = 0 ∴By factor theorem, g(x) is a factor of p(x . Find the value of k, if x–1 is a factor of p(x) in each of the following cases: (i) p(x) = x2+x+k Solution: If x-1 is a factor of p(x), then p(1) = 0 By Factor Theorem ⇒ (1)2+(1)+k = 0 ⇒ 1+1+k = 0 ⇒ 2+k = 0 ⇒ k = −2 (ii) p(x) = 2x2+kx+√2 Solution: If x-1 is a factor of p(x), then p(1)=0 ⇒ 2(1)2+k(1)+√2 = 0 ⇒ 2+k+√2 = 0 ⇒ k = −(2+√2) (iii) p(x) = kx2–√2x+1 Solution: If x-1 is a factor of p(x), then p(1)=0 By Factor Theorem ⇒ k(1)2-√2(1)+1=0 ⇒ k = √2-1 (iv) p(x)=kx2–3x+k Solution: If x-1 is a factor of p(x), then p(1) = 0 By Factor Theorem ⇒ k(1)2–3(1)+k = 0 ⇒ k−3+k = 0 ⇒ 2k−3 = 0 ⇒ k= 3/2 Solution: Let p(x) = x3–2x2–x+2 Factors of 2 are ±1 and ± 2 Now, p(x) = x3–2x2–x+2 p(−1) = (−1)3–2(−1)2–(−1)+2 = −1−2+1+2 = 0 Now, Dividend = Divisor × Quotient + Remainder (x+1)(x2–3x+2) = (x+1)(x2–x–2x+2) = (x+1)(x(x−1)−2(x−1)) = (x+1)(x−1)(x-2) (ii) x3–3x2–9x–5 Solution: Let p(x) = x3–3x2–9x–5 Factors of 5 are ±1 and ±5 By trial method, we find that p(5) = 0 So, (x-5) is factor of p(x) Now, p(x) = x3–3x2–9x–5 p(5) = (5)3–3(5)2–9(5)–5 = 125−75−45−5 = 0 Now, Dividend = Divisor × Quotient + Remainder (x−5)(x2+2x+1) = (x−5)(x2+x+x+1) = (x−5)(x(x+1)+1(x+1)) = (x−5)(x+1)(x+1) (iii) x3+13x2+32x+20 Solution: Let p(x) = x3+13x2+32x+20 Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20 By trial method, we find that p(-1) = 0 So, (x+1) is factor of p(x) Now, p(x)= x3+13x2+32x+20 p(-1) = (−1)3+13(−1)2+32(−1)+20 = −1+13−32+20 = 0