How do you factorize the given polynomial by grouping $8{{r}^{3}}-64{{r}^{2}}+r-8$?

Hint: We start solving the problem by recalling the steps of the method of factorization by grouping. We then group the first two terms of polynomials and the last two terms of polynomials together. We then factor out the GCF (Greatest common factor) for each separate binomial to proceed through the problem. We then need to factor out the GCF (Greatest common factor) of the whole polynomial to get the required answer.

Complete step-by-step answer:
According to the problem, we are asked to factorize the given polynomial $8{{r}^{3}}-64{{r}^{2}}+r-8$ by grouping.
We have given the polynomial $8{{r}^{3}}-64{{r}^{2}}+r-8$.
Let us recall the concept of factor by grouping.
We know that the factor by grouping of polynomial with 4 terms is done as follows:
Step1: We need to group the first two terms together and then the last two terms together.
Step 2: We then need to factor out the GCF (Greatest common factor) for each separate binomial.
Step 3: We then need to factor out the GCF (Greatest common factor) of the whole polynomial.
So, let us group the given polynomial as $8{{r}^{3}}-64{{r}^{2}}+r-8=\left( 8{{r}^{3}}-64{{r}^{2}} \right)+\left( r-8 \right)$ ---(1).
Now, let us factor the GCF (Greatest common factor) of the binomial $8{{r}^{3}}-64{{r}^{2}}$.
So, we have $8{{r}^{3}}-64{{r}^{2}}=\left( r\times 8{{r}^{2}} \right)+\left( -8\times 8{{r}^{2}} \right)$. We can see that the GCF is $8{{r}^{2}}$.
So, the factorization of $8{{r}^{3}}-64{{r}^{2}}$ is $8{{r}^{3}}-64{{r}^{2}}=8{{r}^{2}}\times \left( r-8 \right)$ ---(2).
Now, let us factor the GCF (Greatest common factor) of the binomial $r-8$.
So, we have $r-8=\left( r\times 1 \right)+\left( -8\times 1 \right)$. We can see that the GCF is 1.
So, the factorization of $r-8$ is $r-8=1\times \left( r-8 \right)$ ---(3).
Let us substitute equations (2) and (3) in equation (1).
$\Rightarrow 8{{r}^{3}}-64{{r}^{2}}+r-8=8{{r}^{2}}\times \left( r-8 \right)+1\times \left( r-8 \right)$.
$\Rightarrow 8{{r}^{3}}-64{{r}^{2}}+r-8=\left( 8{{r}^{2}}+1 \right)\left( r-8 \right)$.
So, the factorization of the polynomial $8{{r}^{3}}-64{{r}^{2}}+r-8$ as $\left( 8{{r}^{2}}+1 \right)\left( r-8 \right)$.
$\therefore $ The factorization of the polynomial $8{{r}^{3}}-64{{r}^{2}}+r-8$ is $\left( 8{{r}^{2}}+1 \right)\left( r-8 \right)$.

Note: We should perform each step carefully in order to avoid calculation mistakes and confusion. We can also factorize the given by finding the zeros of the given polynomial. We should know that the term $8{{r}^{2}}+1$ cannot be further reduced to the factors $\left( r-a \right)$, $\left( r-b \right)$ where a, b are real numbers. Similarly, we can expect problems to factorize the polynomial $3{{r}^{3}}-81{{r}^{2}}+r-27$.