Question

How do you factor $4{{m}^{4}}-12{{m}^{2}}+8m$ ?
(a) Factor by grouping
(b) Zero putting
(c) Guessing the factors
(d) None of the above

Answer 1

Hint: To find the factor of the given equation $4{{m}^{4}}-12{{m}^{2}}+8m$, we will try to factorize it by grouping them among terms. We will start off by analyzing the terms and check if there is a common factor between them or not. Then we continue with adding or subtracting the terms and then factorize them to get the needed answer and factorization.

Complete step by step solution:
According to the question, we have our given equation as, $4{{m}^{4}}-12{{m}^{2}}+8m$ and we are to factorize this equation.
So, to start with,
$4{{m}^{4}}-12{{m}^{2}}+8m$
As we can easily see, $4m$ is a common factor present in all the terms, we can take it as a common factor.
$=4m({{m}^{3}}-3m+2)$
And also, we are trying to factorize this with grouping, we will check which one of the terms are similar and in the next step we will take common term from the terms and we also see that putting the value of m as 1 gives us a value 0 of our given term, so $(m-1)$ is a factor of the given equation.
We will now try to set up ${{m}^{3}}-3m+2$ as the factors of $(m-1)$.
${{m}^{3}}-3m+2$
Now, from here, we will try to add and subtract ${{m}^{2}}$to get our needed factor of $(m-1)$. Along with that we can also write -3m as –m -2m to go ahead with the problem.
$\Rightarrow {{m}^{3}}-{{m}^{2}}+{{m}^{2}}-m-2m+2$
Grouping them together and taking common,
$\Rightarrow {{m}^{2}}(m-1)+m(m-1)-2(m-1)$
Now, taking $(m-1)$common from all of them we get,
$\Rightarrow ({{m}^{2}}+m-2)(m-1)$
One has to determine all the terms that were multiplied to obtain the given polynomial. Then try to factor every term that you got in the first step and this continues until you cannot factor further. When you can’t perform any more factoring, it is said that the polynomial is factored completely.
Again, we can factorize ${{m}^{2}}+m-2$,
${{m}^{2}}+m-2$
we can write m as 2m – m also,
$\Rightarrow {{m}^{2}}+2m-m-2$
Taking m common from first two terms and taking -1 common from last two terms, we get,
$\Rightarrow m(m+2)-1(m+2)$
Now, grouping them together,
$\Rightarrow (m+2)(m-1)$
So, now we have,
$4{{m}^{4}}-12{{m}^{2}}+8m=(m+2)(m-1)(m-1)$
Hence the solution is, (a) Factor by grouping

Note: A polynomial $4{{m}^{4}}-12{{m}^{2}}+8m$ can be written as a product of two or more polynomials of degree less than or equal to that of it. Each polynomial involved in the product will be a factor of it. Monomials can be factorized in the same way as integers, just by writing the monomial as the product of its constituent prime factors. In the case of monomials, these prime factors can be integers as well as other monomials which cannot be factored further.