9x2+19x2+19x^2 + \dfrac{1}{9x^2} + 19x2+9x21+1 in the form of factors is :
(3x+13x+1)(3x+13x−1)\Big(3x + \dfrac{1}{3x} + 1\Big)\Big(3x + \dfrac{1}{3x} - 1\Big)(3x+3x1+1)(3x+3x1−1)
(3x−13x+1)(3x−13x−1)\Big(3x - \dfrac{1}{3x} + 1\Big)\Big(3x - \dfrac{1}{3x} - 1\Big)(3x−3x1+1)(3x−3x1−1)
(3x+13x)(3x−13x+1)\Big(3x + \dfrac{1}{3x}\Big)\Big(3x - \dfrac{1}{3x} + 1\Big)(3x+3x1)(3x−3x1+1)
(3x−13x)(3x+13x−1)\Big(3x - \dfrac{1}{3x}\Big)\Big(3x + \dfrac{1}{3x} - 1\Big)(3x−3x1)(3x+3x1−1)
16 Likes
Given,
=9x2+19x2+1=9x2+19x2+2−1=(3x)2+(13x)2+2−1=(3x+13x)2−12=(3x+13x+1)(3x+13x−1).\phantom{=} 9x^2 + \dfrac{1}{9x^2} + 1 \\[1em] = 9x^2 + \dfrac{1}{9x^2} + 2 - 1 \\[1em] = (3x)^2 + \Big(\dfrac{1}{3x}\Big)^2 + 2 - 1 \\[1em] = \Big(3x + \dfrac{1}{3x}\Big)^2 - 1^2 \\[1em] = \Big(3x + \dfrac{1}{3x} + 1\Big)\Big(3x + \dfrac{1}{3x} - 1\Big).=9x2+9x21+1=9x2+9x21+2−1=(3x)2+(3x1)2+2−1=(3x+3x1)2−12=(3x+3x1+1)(3x+3x1−1).
Hence, Option 1 is the correct option.
Answered By
10 Likes
27x3 - 48x in the form of factors is :
3(3x - 4)(3x - 4)
3x(3x + 4)(3x + 4)
3(3x2 - 4x)(3x + 4)
3x(3x - 4)(3x + 4)
x2 - (x - 4y)2 in the form of factors is :
8y(x - 2y)
8y(x + 2y)
4y(x + 2y)
4y(x - 2y)
Factorise :
a2 - (2a + 3b)2
25(2a - b)2 - 81b2
