Question

State whether True or False.
The H.C.F. and L.C.M. of:
\[{a^2}{b^2} - {\text{ }}{b^4},{\text{ }}a{b^2} + {\text{ }}{b^3}and{\text{ }}ab{\text{ }}-{\text{ }}{b^2}is{\text{ }}b;{\text{ }}{b^2}\left( {{a^2} - {\text{ }}{b^2}} \right)\]
If true then enter \[1\] and if false then enter \[0.\]

Answer 1

Hint: We will start by taking factors of \[{a^2}{b^2} - {\text{ }}{b^4},{\text{ }}a{b^2} + {\text{ }}{b^3}and{\text{ }}ab{\text{ }}-{\text{ }}{b^2}is{\text{ }}b;{\text{ }}{b^2}\left( {{a^2} - {\text{ }}{b^2}} \right)\]\[.\]
Then we will obtain H.C.F. and L.C.M. and check whether it’s true or false.

Complete step-by-step answer:
The relation between H.C.F. and L.C.M. of two polynomials is the product of the two polynomials is equal to the product of their H.C.F. and L.C.M. If p(x) and q(x) are two polynomials, then p(x) ∙ q(x) = {H.C.F. of p(x) and q(x)} x {L.C.M. of p(x) and q(x)}. 1.

Step 1: On factorizing \[{a^2}{b^2} - {\text{ }}{b^4},\] we get
\[{a^2}{b^2} - {\text{ }}{b^4} = {\text{ }}{b^2}\left( {{a^2} - {\text{ }}{b^2}} \right){\text{ }} = {\text{ }}{b^2}\left( {{a^{}} + {\text{ }}b} \right){\text{ }}\left( {{a^{}} - {\text{ }}b} \right)\;\; \ldots .\left( 1 \right)\;\]
Step 2: On factorizing \[a{b^2} + {\text{ }}{b^3},\] we get
\[a{b^2} + {\text{ }}{b^3} = {\text{ }}{b^2}\left( {{a^{}} + {\text{ }}b} \right)\;\;\; \ldots .\left( 2 \right)\;\]
Step 3: On factorizing \[ab{\text{ }}-{\text{ }}{b^2},\] we get
\[ab{\text{ }}-{\text{ }}{b^2} = {\text{ }}b{\text{ }}\left( {{a^{}} - {\text{ }}b} \right)\;....\left( 3 \right)\;\]
Now from, \[\left( 1 \right),{\text{ }}\left( 2 \right),{\text{ }}\left( 3 \right)\]
H.C.F. \[ = {\text{ }}b\]
L.C.M. \[ = {\text{ }}{b^2}\left( {{a^2} - {\text{ }}{b^2}} \right)\]
Thus, true H.C.F. and L.C.M. is \[b\] and \[{b^2}\left( {{a^2} - {\text{ }}{b^2}} \right).\]Hence, \[1.\]

Note: H.C.F. stands for highest common factors, is the greatest number which divides the given numbers. L.C.M. stands for Least Common Multiple, is the smallest number which is a multiple of given numbers.