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 : If x+1 is a factor of the polynomial 3x(square) - kx,then find the value of k. -Maths 9th
Last Answer : Solution :-
Description : If the sum of zeroes of the quadratic polynomial 3x2 – kx + 6 is 3, then find the value of k. -Maths 9th
Last Answer : Here a = 3, b = -k, c = 6 Sum of the zeroes, (α + β) = − = 3 …..(given) ⇒ −(−)3 = 3 ⇒ k = 9
Description : For what value of k,(x+1) is a factor of p(x) - kx(square) - x - 4 ? -Maths 9th
Description : One of the zeroes of the polynomial 2x2 + 7x – 4 is -Maths 9th
Last Answer : (b) Let p (x) = 2x2 + 7x-4 = 2x2 + 8x-x-4 [by splitting middle term] = 2x(x+ 4)-1(x+ 4) = (2x-1)(x+ 4) For zeroes of p(x), put p(x) = 0 ⇒ (2x -1) (x + 4) = 0 ⇒ 2x-1 = 0 and x+4 = 0 ⇒ x = 1/2 and x = -4 Hence, one of the zeroes of the polynomial p(x) is 1/2.
Description : One root of x^2 + kx – 8 = 0 is the square of the other, then the value of k is : -Maths 9th
Last Answer : answer:
Description : If the HCF of (x^2 + x – 12) and (2x^2 – kx – 9) is (x – k), then what is the value of k ? -Maths 9th
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 : For what value of k, will the expression (3x^3 – kx^2 + 4x + 16) be divisible by (x – k/2) ? -Maths 9th
Last Answer : Given f(x) = 3x³ - kx² + 4x + 16. Since (x - k/2) is a factor of polynomial. This means x = k/2 is the zero of the given polynomial. ⇒ f(k/2) = 3(k/2)³ - k(k/2)² + 4(k/2) + 16 ⇒ 0 ... - 4k + 32) + 4(k² - 4k + 32) ⇒ 0 = (k + 4)(k² - 4k + 32) ⇒ k = -4.
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 : For what value of k will the roots of the equation kx^2 – 5x + 6 = 0 be in the ratio 2 : 3 ? -Maths 9th
Last Answer : (b) 1 Let the roots of the equation kx2 - 5x + 6 = 0 be α and β. Then, α + β = \(\frac{5}{k}\) ...(i) αβ = \(\frac{6}{k}\) ...(ii) Given \(\ ... frac{9}{k}\) ⇒ 9k2 - 9k = 0 k(k - 1) = 0 ⇒ k = 0 or 1 But k = 0 does not satisfy the condition, so k = 1.
Description : Find the values of k for which the equations x^2 – kx – 21 = 0 and x^2 – 3kx + 35 = 0 will have a common root? -Maths 9th
Last Answer : Let the common root be α ⟹α2−kα−21=0......(1) α2−3kα+35=0........(2) (1)−(2)⟹2kα=56⟹α=k28Substituting α=k28 in (1) (k28)2−28−21=0 ⟹k=±4
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 : 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 : 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.
Description : p(x)=x3-2x2-4x-1, g(x)=x- -Maths 9th
Last Answer : NEED ANSWER
Last Answer : This answer was deleted by our moderators...
Description : factorize x3-2x2-x+2 -Maths 9th
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 : 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 : If (x + k) is a common factor of x^2 + px + q and x^2 + lx + m, then the value of k is -Maths 9th
Description : Show that 2x+1 is a factor of polynomial 2x(cube) - 11x(square) - 4x + 1. -Maths 9th
Description : Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. -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 : 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 : Find the value of polynomial 12x(square) - 7x + 1, when x=1/4. -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 : 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 : 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 : 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 : 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 : 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 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 : 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 : 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 : Check whether polynomial p(x) = 2x(cube) - 9x(square) + x + 12 is a multiple of 2x-3 or not. -Maths 9th
Description : write the degree of the polynomial p(x)=4 -Maths 9th
Last Answer : f(x)=4 =4*x0 So degree =0