Question

How do you factor ${{x}^{5}}+2{{x}^{4}}-3{{x}^{3}}$?

Answer 1

Hint: In this question we will factorize the polynomial by first reducing the polynomial by taking out the term ${{x}^{3}}$ from all the terms in the polynomial. We will then solve the simplified polynomial by considering it as a quadratic equation and then find the factors of the equation, which are multiplication of linear terms that make up the polynomial.

Complete step by step solution:
We have the given equation as:
$\Rightarrow {{x}^{5}}+2{{x}^{4}}-3{{x}^{3}}$
Now since the term ${{x}^{3}}$ is common in all the terms, we will take it out as common, on taking it out as common, we get:
$\Rightarrow {{x}^{3}}\left( {{x}^{2}}+2x-3 \right)$
Now we have the term $\left( {{x}^{2}}+2x-3 \right)$ in the form of a quadratic equation, therefore we solve it by splitting the middle term.
On splitting the middle term of the equation, we get:
$\Rightarrow {{x}^{3}}\left( {{x}^{2}}+3x-x-3 \right)$
Now on taking the common terms, we get:
$\Rightarrow {{x}^{3}}\left( x\left( x+3 \right)-1\left( x+3 \right) \right)$
Now since the term $\left( x+3 \right)$ is common in both the terms, we can take it out as common and write the equation as:
$\Rightarrow {{x}^{3}}\left( x+3 \right)\left( x-1 \right)$
Now since there are no more terms which are non-linear, this is the factorized form of the equation therefore:
${{x}^{5}}+2{{x}^{4}}-3{{x}^{3}}={{x}^{3}}\left( x+3 \right)\left( x-1 \right)$ , which is the required solution.

Note: It is to be remembered that a polynomial equation is a combination of variables and their coefficients, the equation given above is a $5th$ degree polynomial equation.
The factors of a polynomial equation are the terms of degree $1$, which have to be multiplied among themselves so that we get the required polynomial equation. In the question above, $x$, $\left( x+3 \right)$ and $\left( x-1 \right)$ are factors of the given polynomial.
It is not necessary that all the quadratic equations would have roots which are integer numbers or real numbers therefore quadratic formula is used to solve these types of questions, the quadratic formula is:
$({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Where $({{x}_{1}},{{x}_{2}})$ are the roots of the equation and $a,b,c$ are the coefficients of the quadratic equation.