f(x) = x^4 – 2x^3 + 3x^2 – ax + b is a polynomial such that when it is divided by (x – 1) and (x + 1), the remainders are respectively 5 and 19. -Maths 9th

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Description : When f(x) = x4 - 2x3 + 3x2 - ax is divided by x + 1 and x - 1 , we get remainders as 19 and 5 respectively . -Maths 9th

Last Answer : When f(x) is divided by (x+1) and (x-1) , the remainders are 19 and 5 respectively . ∴ f(-1) = 19 and f(1) = 5 ⇒ (-1)4 - 2 (-1)3 + 3(-1)2 - a (-1) + b = 19 ⇒ 1 +2 + 3 + a + b = 19 ∴ a + b = 13 ------- ... + 3x2 - 5x + 8 ⇒ f(3) = 34 - 2 33 + 3 32 - 5 3 + 8 = 81 - 54 + 27 - 15 + 8 = 47

Description : When f(x) = x4 - 2x3 + 3x2 - ax is divided by x + 1 and x - 1 , we get remainders as 19 and 5 respectively . -Maths 9th

Last Answer : When f(x) is divided by (x+1) and (x-1) , the remainders are 19 and 5 respectively . ∴ f(-1) = 19 and f(1) = 5 ⇒ (-1)4 - 2 (-1)3 + 3(-1)2 - a (-1) + b = 19 ⇒ 1 +2 + 3 + a + b = 19 ∴ a + b = 13 ------- ... + 3x2 - 5x + 8 ⇒ f(3) = 34 - 2 33 + 3 32 - 5 3 + 8 = 81 - 54 + 27 - 15 + 8 = 47

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Description : The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Last Answer : p(x) is divided by x+ 2 =

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Last Answer : Given ax^3 + 4x^2 + 3x - 4 and x^3 - 4x + a leave the same remainder when divided by x - 3. Let p(x) = ax^3 + 4x^2 + 3x - 4 and g(x) = x^3 - 4x + a By remainder theorem, if f(x) is divided by (x − a) then ... 4 27a+41 g(3)=27-4(3)+a 15+a f(3)=G(3) 27a+41=15+a 26a=15-41 a=15-41/26 a=-26/26 a=-1

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Description : Zero of the polynomial p(x)=2x+5 is -Maths 9th

Last Answer : (b) Given, p(x) = 2x+5 For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 ⇒ -5/2 Hence, zero of the polynomial p(x) is -5/2.

Description : Zero of the polynomial p(x)=2x+5 is -Maths 9th

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Last Answer : p(x) = x4 - 3x2 + 2x - 5 According to remainder theorem, the required remainder will be = p(2) p(x) = x4 - 3x2 + 2x - 5 ∴ p(2) = 24 - 3(2)2 + 2(2) - 5 =16 - 12 + 4 - 5 = 3

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Last Answer : Actual division method

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Last Answer : p(x) = x4 - 3x2 + 2x - 5 According to remainder theorem, the required remainder will be = p(2) p(x) = x4 - 3x2 + 2x - 5 ∴ p(2) = 24 - 3(2)2 + 2(2) - 5 =16 - 12 + 4 - 5 = 3

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Last Answer : Actual division method

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Last Answer : Answer – B (15110) Explanation – Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10, (108 – 98) = 10 & (140 – 130) = 10. Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10 = 15120 – 10 = 15110

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Last Answer : Let f(x)=px3+x2−2x−q Since f(x) is divisible by (x−1) and (x+1) so x=1 and −1 must make f(x)=0. Therefore, p+1−2−q=0, i.e., p−q=1; and −p+1+2−q=0, i.e., p+q=3 Thus p=2 and q=1

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