Question

Using factor theorem, factorize: ${{x}^{3}}-6{{x}^{2}}+3x+10$

Answer 1

Hint: To start with, we are to find the value of the terms after factoring it. So, we will start with getting one zero of the problem. Once we find it, we will get one factor of the problem. After that, we will check if the result we get can be factorized or not. If it can be factorized then we will factorize it and get the simplest form of the factors.

Complete step by step answer:
According to the problem, using factor theorem, we are to factorize ${{x}^{3}}-6{{x}^{2}}+3x+10$.
Again, we also know, to factorize any term, we need to know at least one zero of that polynomial. Once we know it, we can divide the polynomial by the factor to find the quotient and factor the quotient further to find other zeros.
Let us start with, x = 1. If we get the value of the polynomial as zero, then 1 is a root of the polynomial.
So, we get, putting the value, ${{1}^{3}}-6\times {{1}^{2}}+3\times 1+10=1-6+3+10=8\ne 0$. Then, 1 is not a root.
Again, putting x = 2, we are getting, ${{2}^{3}}-6\times {{2}^{2}}+3\times 2+10=8-24+6+10=0$
Then, we can see, 2 is a zero of the given term.
Thus, x – 2 is a factor.
Now, ${{x}^{3}}-6{{x}^{2}}+3x+10$can be written as,
$\Rightarrow {{x}^{3}}-2{{x}^{2}}-4{{x}^{2}}+8x-5x+10$
Taking proper terms common,
$\Rightarrow {{x}^{2}}\left( x-2 \right)-4x\left( x-2 \right)-5\left( x-2 \right)$
Now, writing them altogether, we get,
$\Rightarrow \left( {{x}^{2}}-4x-5 \right)\left( x-2 \right)$
Again, $\left( {{x}^{2}}-4x-5 \right)$can also be factorized.
Using the middle term factor process,
$\Rightarrow {{x}^{2}}-5x+x-5$
Taking terms common,
$\Rightarrow x\left( x-5 \right)+1\left( x-5 \right)$
So, we get know, $\left( {{x}^{2}}-4x-5 \right)=\left( x+1 \right)\left( x-5 \right)$

Thus, the solution goes, ${{x}^{3}}-6{{x}^{2}}+3x+10=\left( x+1 \right)\left( x-5 \right)\left( x-2 \right)$

Note: A polynomial ${{x}^{3}}-6{{x}^{2}}+3x+10$ 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.