Description : If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th
Last Answer : The value of a
Description : If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th
Last Answer : p(x) = (k2 – 14) x2 – 2x – 12 Here a = k2 – 14, b = -2, c = -12 Sum of the zeroes, (α + β) = 1 …[Given] ⇒ − = 1 ⇒ −(−2)2−14 = 1 ⇒ k2 – 14 = 2 ⇒ k2 = 16 ⇒ k = ±4
Description : Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. -Maths 9th
Last Answer : Solution :-
Description : Simplify: (i) (a + b + c)2 + (a – b + c)2 (ii) (a + b + c)2 – (a – b + c)2 (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2 (iv) (2x + p – c)2 – (2x – p + c)2 (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2 -Maths 9th
Last Answer : answer:
Description : Determine which of the following polynomials has (x + 1) a factor: (i) x3+x2+x+1 -Maths 9th
Last Answer : 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
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 =
Description : Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th
Last Answer : (i) (2x - 1/x)2 [Use identity: (a - b)2 = a2 + b2 - 2ab ] (2x - 1/x)2 = (2x) 2 + (1/x)2 - 2 (2x)(1/x) = 4x2 + 1/x2 - 4 (ii) (2x + y) (2x - y) [Use identity: (a - b)(a + b) = a2 - b 2 ] (2x + y) (2x - ... ) = a2 - b 2 ](1.5 x 2 - 0.3y2 ) (1.5x2 + 0.3y2 ) = (1.5 x 2 ) 2 - (0.3y2 ) 2 = 2.25 x4 - 0.09y4
Description : Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th
Last Answer : Here, p(x) = x4 + 2x3 - 4x2 + 6x - 3, g(x) = x2 - x +1 On dividing p(x) by g(x) Therefore (x-1) must be subtracted from the polynomial p(x) to make it divisible by g(x).
Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th
Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1
Description : Check whether polynomial p(x) = 2x(cube) - 9x(square) + x + 12 is a multiple of 2x-3 or not. -Maths 9th
Description : Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x - 2x = (- 2) (3-x) (iii) (x - 2) (x + 1) = (x - 1) (x + 3) (iv) (x - 3) (2x + 1) = x (x + 5) (v) (2x - 1) (x - 3) = (x ... vi) x2 + 3x + 1 = (x - 2)2 (vii) (x + 2)3 = 2x(x2 - 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th
Last Answer : this is the correct answer!
Description : By remainder theorem, find the remainder when p(x) is divided by g(x) -Maths 9th
Last Answer : Find the remainder when p(x) is divided by g(x)
Description : BY REAMINDER THEOREM FIND THE REAMINDER, WHEN P(x) IS DIVIDED BY G(X), WHERE -Maths 9th
Last Answer : NEED ANSWER
Last Answer : This answer was deleted by our moderators...
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 : Find the value of k if (x-2)is a factor of polynomial p(x) = 2x(cube) - 6x(square) + 5x + k. -Maths 9th
Description : Check whether p(x) is a multiple of g(x) or not -Maths 9th
Last Answer : p(x) is a multiple of g(x) or not
Description : Show that, x + 3 is a factor of 69 + 11c – x2 + x3 -Maths 9th
Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th
Last Answer : Solution of this question
Description : Factorise : 2x3 -3x2 -17x + 30 -Maths 9th
Last Answer : Factorisation of following
Description : Factorise: 2x3 - 3x2 - 17x + 30. -Maths 9th
Description : If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e , show that a + c + e = b + d = 0. -Maths 9th
Last Answer : Since x2 - 1 = (x - 1) is a factor of p(x) = ax4 + bx3 + cx2 + dx + e ∴ p(x) is divisible by (x+1) and (x-1) separately ⇒ p(1) = 0 and p(-1) = 0 p(1) = a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0 ... (b+d) = 0 ⇒ b + d = 0 ---- (iii) comparing equations (ii) and (iii) , we get a + c + e = b + d = 0
Description : Using factor theorem,show that (x-y) is a factor of x(y(square) - z(square)) + y(z(square) - x(square)) + z(x(square) - y(square) ) -Maths 9th
Description : Let g(x) = 2x and h(x) = x2 + 4. Evaluate (h ∘ g)(−5)?
Last Answer : 104
Description : Find the remainder when f(x)=4x(cube) - 12x(square) +14x - 3 is divided by g(x) = (2x-1). -Maths 9th
Last Answer : ____2x2-5x+4________________ 2x-1 ) 4x3-12x2+14x-3( 4x3-2x2 - + ____________ 0 -10x2+14x-3 ... + ___________ X+1
Description : If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3, then find the value of a. -Maths 9th
Last Answer : Let p(x) = a5 -4a2x3 +2x + 2a +3 Since, x + 2a is a factor of p(x), then put p(-2a) = 0 (-2a)5 – 4a2 (-2a)3 + 2(-2a) + 2a + 3 = 0 ⇒ -32a5 + 32a5 -4a + 2a+ 3 = 0 ⇒ -2a + 3 = 0 2a =3 a = 3/2. Hence, the value of a is 3/2.
Description : If (x+1) is a factor of ax(cube) + x(square) - 2x + 4a - 9,find the value of a. -Maths 9th
Last Answer : solution :-
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 : Find the value of the polynomial p(x) = x^3-3x^2-2x+6 at x = underroot 2 -Maths 9th
Last Answer : In this chapter, we shall proceed with recalling some of the constructions already learnt in the earlier classes and deal with some more. Here in this section, we will construct some of these ... be done? 2. Always explain the construction. Write the sequence of steps that are actually taken.
Description : For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th
Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th
Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.
Description : When (x^3 – 2x^2 + px – q) is divided by (x^2 – 2x – 3), the remainder is (x – 6), What are the values of p and q respectively ? -Maths 9th
Description : If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th
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
Description : Determine the ratio in which 2x +3y – 30 = 0 divides the join of A(3, 4) and B(7, 8) and at what point? -Maths 9th
Last Answer : Let A(1, 2) and B(11, 9) be the given points. Let the points of trisection be P and Q. Then,AP = PQ = QB = k (say)⇒ AQ = AP + PQ = 2k and PB = PQ + QB = 2k ∴ AP : PB = k : 2k = 1 : 2 and AQ ... two points of trisection are \(\big(rac{13}{3},rac{13}{3}\big)\) and \(\big(rac{23}{3},rac{20}{3}\big)\).
Description : Write the coefficient of x2 in each of the following -Maths 9th
Last Answer : The coefficient of x2 in each of the following .
Description : Explain Factor Theorem : -Maths 9th
Description : Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th
Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.