Respuesta :

Answer:

(x+3)(3x-5)(3x+5)

Step-by-step explanation:

Let's consider the possible rational zeros:  factors of -75 over factors of 9

So one such possible 0 is -3

let's try it and see if it works:

-3   |    9          27          -25       -75

     |               -27             0         75

       ------------------------------------------

           9          0            -25         0

So x+3 is a factor and we have another there which is 9x^2+0x-25 or 9x^2-25

So far we have this as the factored form (x+3)(9x^2-25)

The second factor is a difference of squares so we can factored this more:

(x+3)(3x-5)(3x+5)

--------You could have also done factored by grouping here:

9x^3+27x^2-25x-75

(9x^3+27x^2)+(-25x-75)

9x^2(x+3)+-25(x+3)

(x+3)(9x^2-25)

(x+3)(3x-5)(3x+5)

dhiab

Answer:

Step-by-step explanation:

9n^3 + 27n^2 – 25n – 75=9n²(n+3)-25(n+3)

                                         = (n+3)(9n²- 25)

but : 9n²- 25 = (3n)² - 5²

by identity : a²-b² =(a-b)(a+b)

now : a =3n  and b = 5 you have : 9n²- 25 = (3n)² - 5² = (3n-5)(3n+5)

Factor the polynomial :

9n^3 + 27n^2 – 25n – 75=(n+3)(3n-5)(3n+5)

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