Factor polynomial $$ x^{5}-125x^{2} $$

$$ x^{5}-125x^{2} = x^{2}( x-5 )( x^{2}+5x+25 ) $$

Step 1 :

After factoring out $ x^{2} $ we have:

$$ x^{5}-125x^{2} = x^{2} ( x^{3}-125 ) $$

Step 2 :

To factor $ x^{3}-125 $ 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 = 5 $ , we have:

$$ x^{3}-125 = ( x-5 ) ( x^{2}+5x+25 ) $$

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:

$$ \color{blue}{ b = 5 } ~ \text{ and } ~ \color{red}{ c = 25 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 5 } $ and multiply to $ \color{red}{ 25 } $.

Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 25 }$.

PRODUCT = 25
1     25	-1     -25
5     5	-5     -5

Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ 5 }$, we conclude the polynomial cannot be factored.

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