Factor polynomial $$ 2x^{5}+9x^{4}+9x^{3}+7x^{2}-47x+20 $$
Answer
Explanation
Step 1 :
The polynomial $ 2x^{5}+9x^{4}+9x^{3}+7x^{2}-47x+20 $ 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 $ 2x^{5}+9x^{4}+9x^{3}+7x^{2}-47x+20 $ 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. )
$$ \frac{ 2x^{5}+9x^{4}+9x^{3}+7x^{2}-47x+20 }{ x-1 } = 2x^{4}+11x^{3}+20x^{2}+27x-20 $$that is:
$$ 2x^{5}+9x^{4}+9x^{3}+7x^{2}-47x+20 = ( x-1 )( 2x^{4}+11x^{3}+20x^{2}+27x-20 ) $$Step 2 :
Use rational root test to get
$$ 2x^{4}+11x^{3}+20x^{2}+27x-20 = ( 2x-1 )( x^{3}+6x^{2}+13x+20 ) $$Step 3 :
Use rational root test to get
$$ x^{3}+6x^{2}+13x+20 = ( x+4 )( x^{2}+2x+5 ) $$Step 4 :
Step 4: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:
$$ \color{blue}{ b = 2 } ~ \text{ and } ~ \color{red}{ c = 5 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 2 } $ and multiply to $ \color{red}{ 5 } $.
Step 5: Find out pairs of numbers with a product of $\color{red}{ c = 5 }$.
| PRODUCT = 5 | |
| 1 5 | -1 -5 |
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 2 }$, we conclude the polynomial cannot be factored.