Question

Find the H.C.F. of $(3m^3-12m^2+21m-18)$ and $(6m^3-30m^2+60m-48)$ by using the division method.

• A. $3(m-2)$
• B. $(m-2)$
• C. $4(m-2)$
• D. $2(m-2)$

Answer 1

The correct option is A $3(m-2)$

$(3m^3-12m^2+21m-18)=3(m^3-4m^2+7m-6)$
$(6m^3-30m^2+60m-48)=6(m^3-5m^2+10m-8)$

Therefore, the common factors of the two expressions are 3 and 6. The H.C.F. of 3 and 6 is 3.

$m^3-4m^2+7m-6)m^3-5m^2+10m-8(1m^3-4m^2+7m-6\underset{–}{-+-+}-m^2+3m-2$

$-m^2+3m-2=-1(m^2-3m+2)$

$m^2-3m+2)m^3-4m^2+7m-6(m-1-m^3+3m^2-2m\underset{–}{+-+}-5m^2+5m-6-m^2+3m-2\underset{–}{+-+}2m-4$

$2m-4=2(m-2)$

Thus, the H.C.F. of $(m^3-4m^2+7m-6)$ and $(m^3-5m^2+10m-8)=(m-2)$

Therefore, the H.C.F. of $(3m^3-12m^2+21m-18)$ and $(6m^3-30m^2+60m-48)=3(m-2)$