Factor polynomial $$ x^{6}-3x^{5}-6x^{4}+10x^{3}+21x^{2}+9x $$
Answer
Explanation
Step 1 :
After factoring out $ x $ we have:
$$ x^{6}-3x^{5}-6x^{4}+10x^{3}+21x^{2}+9x = x ( x^{5}-3x^{4}-6x^{3}+10x^{2}+21x+9 ) $$Step 2 :
The polynomial $ x^{5}-3x^{4}-6x^{3}+10x^{2}+21x+9 $ can be factorized by using the rational root test. To apply this test first we need to find
at least one rational zero.
Polynomial roots calculator can be used to find rational zero.
In this case one rational zero is $ x = -1 $.
This means that we can divide polynomial $ x^{5}-3x^{4}-6x^{3}+10x^{2}+21x+9 $ by $ x+1 $ (the factor theorem)
After division we have: (you can use Synthetic Division Calculator for the step-by-step explanation on how to divide polynomials. )
$$ \frac{ x^{5}-3x^{4}-6x^{3}+10x^{2}+21x+9 }{ x+1 } = x^{4}-4x^{3}-2x^{2}+12x+9 $$that is:
$$ x^{5}-3x^{4}-6x^{3}+10x^{2}+21x+9 = ( x+1 )( x^{4}-4x^{3}-2x^{2}+12x+9 ) $$Step 3 :
Use rational root test to get
$$ x^{4}-4x^{3}-2x^{2}+12x+9 = ( x+1 )( x^{3}-5x^{2}+3x+9 ) $$Step 4 :
Use rational root test to get
$$ x^{3}-5x^{2}+3x+9 = ( x+1 )( x^{2}-6x+9 ) $$Step 5 :
Both the first and third terms are perfect squares.
$$ x^2 = \left( \color{blue}{ x } \right)^2 ~~ \text{and} ~~ 9 = \left( \color{red}{ 3 } \right)^2 $$The middle term ( $ -6x $ ) is two times the product of the terms that are squared.
$$ -6x = - 2 \cdot \color{blue}{x} \cdot \color{red}{3} $$We can conclude that the polynomial $ x^{2}-6x+9 $ is a perfect square trinomial, so we will use the formula below.
$$ A^2 - 2AB + B^2 = (A - B)^2 $$In this example we have $ \color{blue}{ A = x } $ and $ \color{red}{ B = 3 } $ so,
$$ x^{2}-6x+9 = ( \color{blue}{ x } - \color{red}{ 3 } )^2 $$