what is the sum of the polynomials? (7x3 – 4x2) + (2x3 – 4x2) -General Knowledge

what is the sum of the polynomials? (7x3 – 4x2) + (2x3 – 4x2) -General Knowledge
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1 Answer

Answer :

7x^3 - 4x^2 + 2x^3 - 4x^2 = 9x^3-8x^2
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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

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).

Find the zeroes of the of the polynomials p(x) = 4x2 – 12x + 9 -Maths 10th

Description : Find the zeroes of the of the polynomials p(x) = 4x2 – 12x + 9 -Maths 10th

Last Answer : P(x)= 4x² - 12x + 9 4x²- 12x + 9 = 0 4x²- 6x - 6x +9 = 0 2x(2x - 3) - 3(2x - 3) = 0 (2x - 3) (2x - 3) x = 3/2 x = 3/2 x = 3/2 ; 3/2 are the zeros of the given polynomial.

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

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

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

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

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

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

Factorise : 2x3 -3x2 -17x + 30 -Maths 9th

Description : Factorise : 2x3 -3x2 -17x + 30 -Maths 9th

Last Answer : Factorisation of following

The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

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 =

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

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

Factorise : 2x3 -3x2 -17x + 30 -Maths 9th

Description : Factorise : 2x3 -3x2 -17x + 30 -Maths 9th

Last Answer : Factorisation of following

The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

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 =

Factorise: 2x3 - 3x2 - 17x + 30. -Maths 9th

Description : Factorise: 2x3 - 3x2 - 17x + 30. -Maths 9th

Last Answer : Solution :-

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

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.

what is the difference of the two polynomials? (9x2 + 8x) – (2x2 + 3x) -General Knowledge

Description : what is the difference of the two polynomials? (9x2 + 8x) – (2x2 + 3x) -General Knowledge

Last Answer : (9x^2 + 8x) - (2x^2 + 3x) = 7x^2 + 5x.

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 + 5) (x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x(x2 – 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

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!

The value of the polynomial 5x – 4x2 + 3, when x = -1 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

Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

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.

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

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

Last Answer : Solution of this question

The value of the polynomial 5x – 4x2 + 3, when x = -1 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

Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

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.

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

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

Last Answer : Solution of this question

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

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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

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

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

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Description : 3x6=5x9x2=4x2?

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Description : Classify the following polynomials as polynomials in one variable, two variables etc. -Maths 9th

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Description : Determine the degree of each of the following polynomials. -Maths 9th

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Description : Classify the following as a constant, linear, quadratic and cubic polynomials . -Maths 9th

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Description : Find p(0), p( 1) and p(-2) for the following polynomials -Maths 9th

Last Answer : (i) Given, polynomial is p(x) = 10x - 4x2 - 3 On putting x = 0,1 and - 2, respectively in Eq. (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 p(1) = 10 (1) - 4 (1 )2 -3 = 10-4-3= 10-7= 3 and p(-2 ... = (-2 + 2)(-2 -2) =0 (-4) = 0 Hence, the values of p(0),p(1) and p(-2) are respectively,-4,-3 and 0.

If the polynomials az3 +4z2 + 3z-4 and z3-4z + 0 leave the same remainder when divided by z – 3, -Maths 9th

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Last Answer : Let p1(z) = az3 +4z2 + 3z-4 and p2(z) = z3-4z + o When we divide p1(z) by z - 3, then we get the remainder p,(3). Now, p1(3) = a(3)3 + 4(3)2 + 3(3) - 4 = 27a+ 36+ 9-4= 27a+ 41 When we ... to' the question, both the remainders are same. p1(3)= p2(3) 27a+41 = 15+a 27a-a = 15 - 41 . 26a = 26 a = -1

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Description : Determine the degree of each of the following polynomials. -Maths 9th

Last Answer : (i) Degree of polynomial 2x-1 is one, Decause the maximum exponent of x is one. (ii) Degree of polynomial -10 or -10x° is zero, because the exponent of x is zero. (iii) Degree of polynomial x3 - ... iv) Degree of polynomial y3(1-y4) or y3 - y7 is seven, because the maximum exponent of y is seven.

Classify the following as a constant, linear, quadratic and cubic polynomials . -Maths 9th

Description : Classify the following as a constant, linear, quadratic and cubic polynomials . -Maths 9th

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Description : Find p(0), p( 1) and p(-2) for the following polynomials -Maths 9th

Last Answer : (i) Given, polynomial is p(x) = 10x - 4x2 - 3 On putting x = 0,1 and - 2, respectively in Eq. (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 p(1) = 10 (1) - 4 (1 )2 -3 = 10-4-3= 10-7= 3 and p(-2 ... = (-2 + 2)(-2 -2) =0 (-4) = 0 Hence, the values of p(0),p(1) and p(-2) are respectively,-4,-3 and 0.

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If the polynomials az3 + 42z2 + 3z – 4 and z3 - 4z + a leave the same remainder when divided by z – 3, find the value of a. -Maths 9th

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Description : NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.1 -Maths 9th

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NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.2 -Maths 9th

Last Answer : 1. Visualise 3.765 on the number line, using successive magnification. 3.765 lies between 3 and 4. Let us divide the interval (3, 4) into 10 equal parts. Since, 3.765 lies between 3.7 and 3. ... 10 equal parts. Now, the point corresponding to 3.765 is clearly located, as shown in Fig. (iii) above.