Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1 -Maths 9th

1 Answer

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 is -3. Similarly, for x = –1, f(–1) = 5(–1) – 4(–1)2 + 3 => f(–1) = –5 –4 + 3 = -6 The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.

Related questions

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 : 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 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 k if (x-2)is a factor of polynomial p(x) = 2x(cube) - 6x(square) + 5x + k. -Maths 9th

Last Answer : Solution :-

Description : Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Last Answer : (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Here, the highest power of x is 4, Hence, the degree of a polynomial is 4.

Description : Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Last Answer : (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Here, the highest power of x is 4, Hence, the degree of a polynomial is 4.

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 : Find the value of 4x2 + y2 + 25z2 + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2. -Maths 9th

Last Answer : 4x²+y²+25z²+4xy-10yz-20zx when x=4, y=3 &z=2 so =>4(4)²+(3)²+ 25(2)²+4(4)(3)-10(3)(2)-20(2)(4) =>64+9+100+48-60-160 =>221-220 =>1

Description : if 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2. -Maths 9th

Last Answer : Given 2x+3y=8 and xy=2, formula, (a+b)2=a2+b2+2ab ∴(2x+3y)2=4x2+9y2+2(2x)(3y) (2x+3y)2=4x2+9y2+12xy 82=4x2+9y2+12(2) ∴4x2+9y2=64−24=40

Description : If ̳α ‘ and ̳β ‘ are the zeroes of a quadratic polynomial x 2 − 5x + b and α − β = 1, then the value of ̳b‘ is (a) – 5 (b) 6 (c) 5 (d) – 6

Last Answer : (b) 6

Description : Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x2–3x+7 -Maths 9th

Last Answer : Solution: The equation 4x2–3x+7 can be written as 4x2–3x1+7x0 Since x is the only variable in the given equation and the powers of x (i.e., 2, 1 and 0) are whole numbers, we can say that the expression 4x2–3x+7 is a polynomial in one variable

Description : Find (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). -Maths 9th

Last Answer : Solution of this question

Description : Find (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). -Maths 9th

Last Answer : Solution of this question

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 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 : 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.

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : The value of 'k' is 4

Description : Find the value of polynomial 12x(square) - 7x + 1, when x=1/4. -Maths 9th

Last Answer : Solution :-

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 : 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 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 : 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 : Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Last Answer : Given, equation is 5x + 2ky = 3k. On putting x =1 and y =1 in this equation, we get 5(1) + 2k(1) =3k ⇒ 5 + 2k =3k ⇒ 5 = 3k - 2k ⇒ k = 5 Hence, required value of k is 5.

Description : Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Last Answer : Given, equation is 5x + 2ky = 3k. On putting x =1 and y =1 in this equation, we get 5(1) + 2k(1) =3k ⇒ 5 + 2k =3k ⇒ 5 = 3k - 2k ⇒ k = 5 Hence, required value of k is 5.

Description : If x = 0 and y = k is a solution of the equation 5x - 3 y = 0, find the value of k. -Maths 9th

Last Answer : Solution :-

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 : If x^3 + 5x^2 + 10k leaves remainder – 2x when divided by x^2 + 2, then what is the value of k ? -Maths 9th

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

Description : If the product of two zeroes of polynomial 2x3 + 3x2 – 5x – 6 is 3, then find its third zero. -Maths 10th

Last Answer : The third zero is 1. Step-by-step explanation: Given the polynomial The product of two zeroes is 3 we have to find the third zero As product of two zeroes is 3 Let ab=3 Therefore, 3c=3 ⇒ c=1 hence, the third zero is 1.

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 : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

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 : 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 : 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 : 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 : 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 : 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 : 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 : 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 : 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 : 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 : 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

Last Answer : Solution :-

Description : write the degree of the polynomial p(x)=4 -Maths 9th

Last Answer : f(x)=4 =4*x0 So degree =0

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: