Question

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

Answer 1

Hint: As the given polynomial is of degree 3 so it is difficult to factor it by using a method like the split middle term or completing the square method. So, first, we will take common terms out from the given polynomial by grouping the terms. Then we will further simplify the obtained equation to get the factors of the given polynomial.

Complete step by step answer:
We have been given a polynomial $ {{x}^{3}}+4{{x}^{2}}+5x+2 $ .
We have to factorize the given polynomial.
Now, we know that there is no suitable method to factorize the polynomial of degree 3. So we need to solve it numerically. Let us assume that $ -1 $ is a root of the equation so $ x+1 $ is the one factor of the given equation. So, by taking $ x+1 $ out as common factor we will get
 $ \begin{align}
  & \Rightarrow {{x}^{3}}+4{{x}^{2}}+5x+2 \\
 & \Rightarrow \left( x+1 \right)\left( {{x}^{2}}+3x+2 \right) \\
\end{align} $
Now, we can factorize the equation $ \left( {{x}^{2}}+3x+2 \right) $ by using a method like splitting the middle term then we will get
 $ \begin{align}
  & \Rightarrow \left( {{x}^{2}}+3x+2 \right) \\
 & \Rightarrow {{x}^{2}}+2x+x+2 \\
\end{align} $
Now, taking the common terms out we will get
 $ \Rightarrow x\left( x+2 \right)+1\left( x+2 \right) $
Now taking the common terms out we will get
 $ \Rightarrow \left( x+2 \right)\left( x+1 \right) $
Now, substituting the factors in the above obtained equation we will get
 $ \Rightarrow \left( x+1 \right)\left( x+1 \right)+\left( x+2 \right) $
Or we can write the factors as $ \left( x+2 \right){{\left( x+1 \right)}^{2}} $.

Note:
Here in this question, we have given a third-degree polynomial. We can easily solve a second-degree polynomial by using any suitable method but solving a third-degree polynomial is difficult. So we need to solve it numerically or try to solve it by using numbers by chance. We can verify the $ \left( x+1 \right)\left( {{x}^{2}}+3x+2 \right) $ equation by multiplying the brackets whether the original equation is formed or not.