Question

If $u\left(x\right)=(x-1{)}^2$ and $v\left(x\right)=(x^2,-1)$ then verify $LCM\times HCF=u\left(x\right)\times v\left(x\right)$.

Answer 1

The given polynomial $v\left(x\right)=x^2-1$ can be rewritten as $v\left(x\right)=(x+1)(x-1)$ because one of the identity is $a^2-b^2=(a+b)(a-b)$.

Now, the LCM and HCF of the given polynomials $u\left(x\right)=(x-1{)}^2$ and $v\left(x\right)=(x+1)(x-1)$ is as follows:

$u\left(x\right)=(x-1{)}^2=(x-1\left)\right(x-1\left)v\right(x)=(x+1\left)\right(x-1)\Rightarrow LCM(u\left(x\right),v\left(x\right))=(x-1\left)\right(x-1\left)\right(x+1)=(x-1{)}^2(x+1)\Rightarrow HCF\left(u\right(x),v(x\left)\right)=(x-1)$

Now, we consider $LCM\times HCF=u\left(x\right)\times v\left(x\right)$ as shown below:

$LCM\times HCF=u\left(x\right)\times v\left(x\right)\Rightarrow (x-1{)}^2(x+1)\times (x-1)=(x-1{)}^2(x+1)(x-1)\Rightarrow (x-1{)}^3(x+1)=(x-1{)}^3(x+1)\Rightarrow LHS=RHS$

Hence, $LCM\times HCF=u\left(x\right)\times v\left(x\right)$ is verified.