Question

Find the square root of (6x2+x-1) (3x2+2x-1) (2x2+3x+1)

Answer 1

Answer:

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Answer 2

Answer:

The square root of (6x²+x-1)(3x²+2x-1)(2x²+3x+1) is 6x³ + 7x² - 1

Step-by-step explanation:

We need to find the square root of: (6x²+x-1)(3x²+2x-1)(2x²+3x+1)

We can write this as below by mid term splitting

= (6x²+3x-2x-1)(3x²+3x-x-1)(2x²+2x+x+1)

= [3x(2x+1)-1(2x+1)] [3x(x+1)-1(x+1)] [(2x(x+1)+1(x+1)]

on factorizing, we get

= (2x+1)(3x-1) (3x-1)(x+1) (x+1)(2x+1)

= [(2x+1)(3x-1)(x+1)]²

We need to find the square root of

(6x²+x-1)(3x²+2x-1)(2x²+3x+1) = [(2x+1)(3x-1)(x+1)]²

=> √[(2x+1)(3x-1)(x+1)]² = (2x+1)(3x-1)(x+1)

On expansion, we get

(6x²+x-1)(x+1)

= 6x³ + 6x² + x² + x - x - 1

= 6x³ + 7x² - 1

Therefore the square root of (6x²+x-1)(3x²+2x-1)(2x²+3x+1) is 6x³ + 7x² - 1