Question

find the HCF and LCM of:
2x³–5x²–3x and x²–9​

Answer 1

Answer:

x

2

−9)=(x−3)(x+3)

(x

3

−27)=(x

3

−3

3

)=(x−3)(x

2

−3x+9)

(x

2

−8x+15)=(x−3)(x−5)

∴ H.C.F. of (x

2

−9),(x

3

−27) and (x

2

−8x+15) is x−3.