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 k if (x-2)is a factor of polynomial p(x) = 2x(cube) - 6x(square) + 5x + k. -Maths 9th
Last Answer : Solution :-
Description : Check whether polynomial p(x) = 2x(cube) - 9x(square) + x + 12 is a multiple of 2x-3 or not. -Maths 9th
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 : 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 : 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
Last Answer : answer:
Description : Show that 2x+1 is a factor of polynomial 2x(cube) - 11x(square) - 4x + 1. -Maths 9th
Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th
Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10
Description : Determine the remainder when polynomial p(x) is divided by x - 2 . -Maths 9th
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
Description : Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. -Maths 9th
Last Answer : Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 For zeroes of polynomial, put p(x) = 0 (x – 2)2 – (x+ 2)2 = 0 (x-2 + x+2)(x-2-x-2) = 0 [using identity, a2-b2 =(a-b)(a + b)] ⇒ (2x)(-4) = 0
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 : write the degree of the polynomial p(x)=4 -Maths 9th
Last Answer : f(x)=4 =4*x0 So degree =0
Description : Degree of the zero polynomial is -Maths 9th
Last Answer : (d) The degree of zero polynomial is not defined, because in zero polynomial, the coefficient of any variable is zero i.e., Ox2 or Ox5,etc. Hence, we cannot exactly determine the degree of variable.
Description : Zero of the zero polynomial is -Maths 9th
Last Answer : (c) Zero of the zero polynomial is any real number. e.g., Let us consider zero polynomial be 0(x-k), where k is a real number For determining the zero, put x-k = 0 ⇒ x = k Hence, zero of the zero polynomial be any real number.
Description : 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 -Maths 9th
Last Answer : 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
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 : Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th
Last Answer : Let p(x) =3x3 – 4x2 + 7x – 5 At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3)- 5 = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.
Description : A cubic polynomial f(x) is such that f(1) = 1, f(2) = 2, f(3) = 3 and f(4) = 5, then f(6) equals : -Maths 9th
Description : If the polynomial x^19 + x^17 + x^13 + x^11 + x^7 + x^5 + x^3 is divided by (x^2 + 1), then the remainder is : -Maths 9th
Description : The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th
Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6
Description : If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th
Last Answer : (c) Let p(x) = 2x2 + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 ⇒ 2-k = 0 ⇒ k= 2 Hence, the value of k is 2.
Description : x + 1 is a factor of the polynomial -Maths 9th
Last Answer : (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. So, x = -1 is zero of x3 + x2 + x+1 (-1)3 + (-1)2 + (-1) + 1 = 0 ⇒ -1+1-1 + 1 = 0 ⇒ 0 = 0 Hence, our assumption is true.
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 : Determine which of the following polynomial has x – 2 a factor -Maths 9th
Last Answer : first option is the correct answer for the given question solution is as follows:- let x-2=0 then, x=2 put x in (i) 3(2)(2)+6(2)-24=0 12+12-24=0 {use BODMAS rule for solution}... 24-24=0 0=0 this verifies our answer
Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th
Last Answer : As the given polynomial divisible by x-2 means the polynomial satisfies for the value x=2 So putting x=2 in x²+(4-k)x+2 yields 0 ⇒2²+(4-k)2+2=0 ⇒4+8-2k+2=0 ⇒ 2k=14 ⇒ k= ... ;-3x+2 if factorized yields (x-1)(x-2). Thus is divisible by x-2 as well as divisible by x-1.
Last Answer : The value of 'k' is 4
Description : Find the value of polynomial 12x(square) - 7x + 1, when x=1/4. -Maths 9th
Description : If x+1 is a factor of the polynomial 3x(square) - kx,then find the value of k. -Maths 9th
Description : If the remainder of the polynomial a0 + a1x + a2x^2 + ....... + anx^n when divided by (x – 1) is 1, then which one of the following is correct ? -Maths 9th
Description : When a polynomial f(x) is divided by (x – 3) and (x + 6), the respective remainders are 7 and 22. What is the remainder when f(x) is -Maths 9th
Description : Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1 -Maths 9th
Last Answer : Let the polynomial be f(x) = 5x – 4x2 + 3 Now, for x = 2, f(2) = 5(2) – 4(2)2 + 3 => f(2) = 10 – 16 + 3 = –3 Or, the value of the polynomial 5x – 4x2 + 3 at x = 2 ... (–1) = –5 –4 + 3 = -6 The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.
Description : SIMplify (2x+p-q)raise to power 2 -(2x-p+q)raise to power2 -Maths 9th
Last Answer : 2p-2q+2 is answer of the question
Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th
Last Answer : x = 1 y = -2 2x-y = p Therefore, p = 2(1)-(-2) = 2 + 2 = 4
Last Answer : 2x-y=p put x-1,y=-2 =2(1)-(-2)=p p=4
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