$(12 m^{2} - 27) = ? (a) (2 m - 3) (3 m - 9) (b) 3 (2 m - 9) (3 m - 1) (c) 3 (2 m - 3) (2 m + 3) (d) n o n e o f t h e s e$

Open in App

Solution

$12 m^{2} - 27 = 3 (4 m^{2} - 9) = 3 {(2 m)^{2} - (3)^{2}} = 3 (2 m - 3) (2 m + 3)$

Suggest Corrections

Similar questions

(2 m + 3) (2 m + 3)

Calculate the value of centripetal acceleration for a mass of $10 k g$ performing a circular motion with $4 m / s$ with the radius $0.5 m$ .

2^{m} \times 3^{m}

16 m n + 8 + 4 n + 32 m

Join BYJU'S Learning Program