Factor polynomial $$ x^{5}-4x^{3}-x^{2}+4 $$
Answer
Explanation
Step 1 :
To factor $ x^{5}-4x^{3}-x^{2}+4 $ we can use factoring by grouping:
Group $ \color{blue}{ x^{5} }$ with $ \color{blue}{ -4x^{3} }$ and $ \color{red}{ -x^{2} }$ with $ \color{red}{ 4 }$ then factor each group.
$$ \begin{aligned} x^{5}-4x^{3}-x^{2}+4 = ( \color{blue}{ x^{5}-4x^{3} } ) + ( \color{red}{ -x^{2}+4 }) &= \\ &= \color{blue}{ x^{3}( x^{2}-4 )} + \color{red}{ -1( x^{2}-4 ) } = \\ &= (x^{3}-1)(x^{2}-4) \end{aligned} $$Step 2 :
To factor $ x^{3}-1 $ we can use difference of cubes formula:
$$ I^3 - II^3 = (I - II)(I^2 + I \cdot II + II^2) $$After putting $ I = x $ and $ II = 1 $ , we have:
$$ x^{3}-1 = ( x-1 ) ( x^{2}+x+1 ) $$Step 3 :
Step 3: 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:
$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.
Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.
| PRODUCT = 1 | |
| 1 1 | -1 -1 |
Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.
Step 4 :
Rewrite $ x^{2}-4 $ as:
$$ x^{2}-4 = (x)^2 - (2)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = x $ and $ II = 2 $ , we have:
$$ x^{2}-4 = (x)^2 - (2)^2 = ( x-2 ) ( x+2 ) $$Step 5 :
Step 5: 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:
$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.
Step 6: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.
| PRODUCT = 1 | |
| 1 1 | -1 -1 |
Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.