BY REAMINDER THEOREM FIND THE REAMINDER, WHEN P(x) IS DIVIDED BY G(X), WHERE -Maths 9th

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Description : BY REAMINDER THEOREM FIND THE REAMINDER, WHEN P(x) IS DIVIDED BY G(X), WHERE -Maths 9th

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Description : p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Last Answer : 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

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Description : A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

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Description : If x51 + 51 is divided by x + 1, then the remainder is -Maths 9th

Last Answer : (d) Let p(x) = x51 + 51 . …(i) When we divide p(x) by x+1, we get the remainder p(-1) On putting x= -1 in Eq. (i), we get p(-1) = (-1)51 + 51 = -1 + 51 = 50 Hence, the remainder is 50.

Description : By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial x4 + 1 and x-1. -Maths 9th

Last Answer : Actual division method

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 : If x51 + 51 is divided by x + 1, then the remainder is -Maths 9th

Last Answer : (d) Let p(x) = x51 + 51 . …(i) When we divide p(x) by x+1, we get the remainder p(-1) On putting x= -1 in Eq. (i), we get p(-1) = (-1)51 + 51 = -1 + 51 = 50 Hence, the remainder is 50.

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Last Answer : x3+5x2+10k =(x2+2)(x+5)+10k−2x−10 ⇒10k−2x−10=−2x ⇒10k−10=0 or k=1.

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Description : If the expression ax^2 + bx + c is equal to 4, when x = 0, leaves a remainder 4 when divided by x + 1 and leaves a -Maths 9th

Last Answer : Given exp. f(x) = ax2 + bx + c ∴ When x = 0, a.0 + b.0 + c = 4 ⇒ c = 4. The remainders when f(x) is divided by (x + 1) and (x + 2) respectively are f(–1) and f(–2). ∴ f( ... 2b = 2 ...(ii) Solving (i) and (ii) simultaneously we get, a = 1, b = 1.