Factor polynomial $$ 250x^{4}+128x $$

$$ 250x^{4}+128x = 2x( 5x+4 )( 25x^{2}-20x+16 ) $$

Step 1 :

After factoring out $ 2x $ we have:

$$ 250x^{4}+128x = 2x ( 125x^{3}+64 ) $$

Step 2 :

To factor $ 125x^{3}+64 $ we can use sum of cubes formula:

$$ I^3 - II^3 = (I + II)(I^2 - I \cdot II + II^2) $$

After putting $ I = 5x $ and $ II = 4 $ , we have:

$$ 125x^{3}+64 = ( 5x+4 ) ( 25x^{2}-20x+16 ) $$

Step 3 :

Step 3: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 25 , b = -20 ~ \text{ and } ~ c = 16 $$

Step 4: Multiply the leading coefficient $\color{blue}{ a = 25 }$ by the constant term $\color{blue}{c = 16} $.

$$ a \cdot c = 400 $$

Step 5: Find out two numbers that multiply to $ a \cdot c = 400 $ and add to $ b = -20 $.

Step 6: All pairs of numbers with a product of $ 400 $ are:

PRODUCT = 400
1     400	-1     -400
2     200	-2     -200
4     100	-4     -100
5     80	-5     -80
8     50	-8     -50
10     40	-10     -40
16     25	-16     -25
20     20	-20     -20

Step 7: Find out which factor pair sums up to $\color{blue}{ b = -20 }$

Step 8: Because none of these pairs will give us a sum of $ \color{blue}{ -20 }$, we conclude the polynomial cannot be factored.

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