factorize x3-2x2-x+2 -Maths 9th

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

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

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Description : Hey guys, does anyone know the factored form of x3-2x2-x+2?

Last Answer : Welcome to Fluther. Two things: 1. We don't do homework problems or provide answers. If you ask a question of the how do I ? variety, we are much more accommodating. That's just how we roll. (And it ... in a way that'll help you to understand it or get credit for it. But the answer is out there.

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 : 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 : 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 : 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 : Factorize this question -Maths 9th

Last Answer : (i) 9x2 - y2 = (3x)2 - (y)2 = (3x + y) (3x - y) (ii) (3 - x)2 - 36x2 = (3 - x)2 - (6x)2 = (3 - x + 6x) (3 - x - 6x) = (3 + 5x) (3 - 7x) (iii) (2x - 3y)2 - (3y + 4y)2 = (2x - 3y + ... + 7y) (iv) 16x4 - y4 = (4x2)2 - (y2)2 = (4x2 + y2) (4x2 - y2) = (4x2 + y2) (2x + y) (2x - y)

Description : Factorize this question -Maths 9th

Last Answer : (i) 9x2 - y2 = (3x)2 - (y)2 = (3x + y) (3x - y) (ii) (3 - x)2 - 36x2 = (3 - x)2 - (6x)2 = (3 - x + 6x) (3 - x - 6x) = (3 + 5x) (3 - 7x) (iii) (2x - 3y)2 - (3y + 4y)2 = (2x - 3y + ... + 7y) (iv) 16x4 - y4 = (4x2)2 - (y2)2 = (4x2 + y2) (4x2 - y2) = (4x2 + y2) (2x + y) (2x - y)

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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 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 : p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Last Answer : p(x) = x3+3x2+3x+1, g(x) = x+2 g(x) = 0 ⇒ x+2 = 0 ⇒ x = −2 ∴ Zero of g(x) is -2. Now, p(−2) = (−2)3+3(−2)2+3(−2)+1 = −8+12−6+1 = −1 ≠ 0 ∴By factor theorem, g(x) is not a factor of p(x

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