If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th
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If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

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Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. -Maths 9th

Description : Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. -Maths 9th

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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

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

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

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

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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

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IF p(x)=10x-4x -Maths 9th

Description : IF p(x)=10x-4x -Maths 9th

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p(x)=x3-2x2-4x-1, g(x)=x- -Maths 9th

Description : p(x)=x3-2x2-4x-1, g(x)=x- -Maths 9th

Last Answer : This answer was deleted by our moderators...

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

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If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th

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

Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Description : Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

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In the linear equation y = 4x + 13, if x is the number of hours a labourer is on work and y are his wages in rupees then draw the graph. Also find the wages when work is done for 6 hours. -Maths 9th

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Last Answer : when the work is done for 6 hours x=6 y=4(6)+13 y=24+13 y=37 the labourer gets Rs.37 if he works for 6hrs

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Description : If x^2 – 4x + 1 = 0, then what is the value of x^3 + 1/x^3? -Maths 9th

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Determine which of the following polynomials has (x + 1) a factor: (i) x3+x2+x+1 -Maths 9th

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

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

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

Show that, x + 3 is a factor of 69 + 11c – x2 + x3 -Maths 9th

Description : Show that, x + 3 is a factor of 69 + 11c – x2 + x3 -Maths 9th

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Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -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.

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

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

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If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

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

Show that, x + 3 is a factor of 69 + 11c – x2 + x3 -Maths 9th

Description : Show that, x + 3 is a factor of 69 + 11c – x2 + x3 -Maths 9th

Last Answer : This answer was deleted by our moderators...

Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -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.

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

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

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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.

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

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For what value of m will the expression 3x^3 + mx^2 + 4x – 4m be divisible by x + 2 ? -Maths 9th

Description : For what value of m will the expression 3x^3 + mx^2 + 4x – 4m be divisible by x + 2 ? -Maths 9th

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For what value of k, will the expression (3x^3 – kx^2 + 4x + 16) be divisible by (x – k/2) ? -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.

If the polynomials ax^3 + 4x^2 + 3x – 4 and x^3 – 4x + a leave the same remainder when divided by (x – 3), the value of a is : -Maths 9th

Description : If the polynomials ax^3 + 4x^2 + 3x – 4 and x^3 – 4x + a leave the same remainder when divided by (x – 3), the value of a is : -Maths 9th

Last Answer : Given ax^3 + 4x^2 + 3x - 4 and x^3 - 4x + a leave the same remainder when divided by x - 3. Let p(x) = ax^3 + 4x^2 + 3x - 4 and g(x) = x^3 - 4x + a By remainder theorem, if f(x) is divided by (x − a) then ... 4 27a+41 g(3)=27-4(3)+a 15+a f(3)=G(3) 27a+41=15+a 26a=15-41 a=15-41/26 a=-26/26 a=-1

When x^3 + 2x^2 + 4x + b is divided by (x + 1), the quotient is x^2 + ax + 3 and the remainder -Maths 9th

Description : When x^3 + 2x^2 + 4x + b is divided by (x + 1), the quotient is x^2 + ax + 3 and the remainder -Maths 9th

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Write the coefficient of x2 in each of the following -Maths 9th

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If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e , show that a + c + e = b + d = 0. -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

Write the coefficient of x2 in each of the following -Maths 9th

Description : Write the coefficient of x2 in each of the following -Maths 9th

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Factorise : x2 + 9x +18 -Maths 9th

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If x2 + 1/x2 = 34, find x3 + 1/x3 - 9. -Maths 9th

Description : If x2 + 1/x2 = 34, find x3 + 1/x3 - 9. -Maths 9th

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the product (x2 – 1) (x4 + x2 + 1) is equal to -Maths 9th

Description : the product (x2 – 1) (x4 + x2 + 1) is equal to -Maths 9th

Last Answer : (x^2-1)(x^4+x^2+1)=x^6+x^4+x^2-x^4-x^2-1. =x^6+1.