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 (example is 28n^4+16n^3-80n^2)

Guest Sep 24, 2017
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28n^4+16n^3-80n^2

 

\(28n^4+16n^3-80n^2\\ =4n^2(7n^2+4n-20)\)

 

I could use the quadrative formula to solve it but I will see if I can figure it out another way.

 

I need 2 numbers that multiply to 7*-20  and add to give 4

One will need to be positive and the other negative and since 4 is positive, the bigger number will be positive.

 

7*-20 = -7*2*2*5 = 14* -10  that works, the numbers are  14 and -10

change 4n into  14n-10n

 

\(=4n^2(7n^2+4n-20)\\ =4n^2(7n^2+14n-10n-20)\\ \text{Now factor in pairs}\\ =4n^2[7n(n+2)-10(n+2)]\\ =4n^2(7n-10)(n+2)\\ \)

Melody  Sep 24, 2017

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