Factor polynomial $$ x^{5}-4x^{3}+8x^{2}-32 $$

$$ x^{5}-4x^{3}+8x^{2}-32 = ( x+2 )^{ 2 }( x^{2}-2x+4 )( x-2 ) $$

Step 1 :

To factor $ x^{5}-4x^{3}+8x^{2}-32 $ we can use factoring by grouping:

Group $ \color{blue}{ x^{5} }$ with $ \color{blue}{ -4x^{3} }$ and $ \color{red}{ 8x^{2} }$ with $ \color{red}{ -32 }$ then factor each group.

$$ \begin{aligned} x^{5}-4x^{3}+8x^{2}-32 = ( \color{blue}{ x^{5}-4x^{3} } ) + ( \color{red}{ 8x^{2}-32 }) &= \\ &= \color{blue}{ x^{3}( x^{2}-4 )} + \color{red}{ 8( x^{2}-4 ) } = \\ &= (x^{3}+8)(x^{2}-4) \end{aligned} $$

Step 2 :

To factor $ x^{3}+8 $ 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 = 2 $ , we have:

$$ x^{3}+8 = ( x+2 ) ( x^{2}-2x+4 ) $$

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:

$$ \color{blue}{ b = -2 } ~ \text{ and } ~ \color{red}{ c = 4 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -2 } $ and multiply to $ \color{red}{ 4 } $.

Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 4 }$.

PRODUCT = 4
1     4	-1     -4
2     2	-2     -2

Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ -2 }$, 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:

$$ \color{blue}{ b = -2 } ~ \text{ and } ~ \color{red}{ c = 4 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -2 } $ and multiply to $ \color{red}{ 4 } $.

Step 6: Find out pairs of numbers with a product of $\color{red}{ c = 4 }$.

PRODUCT = 4
1     4	-1     -4
2     2	-2     -2

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -2 }$, we conclude the polynomial cannot be factored.

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