Step 1 :

To factor $ x^{10}-x^{6}-x^{4}+1 $ we can use factoring by grouping:

Group $ \color{blue}{ x^{10} }$ with $ \color{blue}{ -x^{6} }$ and $ \color{red}{ -x^{4} }$ with $ \color{red}{ 1 }$ then factor each group.

$$ \begin{aligned} x^{10}-x^{6}-x^{4}+1 = ( \color{blue}{ x^{10}-x^{6} } ) + ( \color{red}{ -x^{4}+1 }) &= \\ &= \color{blue}{ x^{6}( x^{4}-1 )} + \color{red}{ -1( x^{4}-1 ) } = \\ &= (x^{6}-1)(x^{4}-1) \end{aligned} $$

Step 2 :

Rewrite $ x^{6}-1 $ as:

$$ x^{6}-1 = (x^{3})^2 - (1)^2 $$

Now we can apply the difference of squares formula.

$$ I^2 - II^2 = (I - II)(I + II) $$

After putting $ I = x^{3} $ and $ II = 1 $ , we have:

$$ x^{6}-1 = (x^{3})^2 - (1)^2 = ( x^{3}-1 ) ( x^{3}+1 ) $$

Step 3 :

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 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:

$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.

Step 5: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.

Step 5 :

To factor $ x^{3}+1 $ we can use sum 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 6 :

Step 6: 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 7: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 8: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.

Step 7 :

Step 7: 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 8: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 9: Because none of these pairs will give us a sum of $ \color{blue}{ -1 }$, we conclude the polynomial cannot be factored.

Step 8 :

Rewrite $ x^{4}-1 $ as:

$$ x^{4}-1 = (x^{2})^2 - (1)^2 $$

Now we can apply the difference of squares formula.

$$ I^2 - II^2 = (I - II)(I + II) $$

After putting $ I = x^{2} $ and $ II = 1 $ , we have:

$$ x^{4}-1 = (x^{2})^2 - (1)^2 = ( x^{2}-1 ) ( x^{2}+1 ) $$

Step 9 :

Step 9: 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 10: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 11: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.

Step 10 :

Step 10: 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 11: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 12: Because none of these pairs will give us a sum of $ \color{blue}{ -1 }$, we conclude the polynomial cannot be factored.

Step 11 :

Rewrite $ x^{2}-1 $ as:

$$ x^{2}-1 = (x)^2 - (1)^2 $$

Now we can apply the difference of squares formula.

$$ I^2 - II^2 = (I - II)(I + II) $$

After putting $ I = x $ and $ II = 1 $ , we have:

$$ x^{2}-1 = (x)^2 - (1)^2 = ( x-1 ) ( x+1 ) $$

Step 12 :

Step 12: 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 13: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 14: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.

Step 13 :

Step 13: 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 14: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 15: Because none of these pairs will give us a sum of $ \color{blue}{ -1 }$, we conclude the polynomial cannot be factored.

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