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Q1)

 

 

Q2)

Q3)

 

thank you!!

Guest Oct 24, 2018
 #1
avatar+90969 
+1

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

 

Factor by grouping

 

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

 

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

 

(x + 5) (x - 5) ( x + 3)    are the binomial factors

 

 

For the second one we have

 

x^4 + 5x^3 -x^2  - 5x      factor by grouping

 

x^3 ( x + 5) - x (x + 5)

 

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

 

x ( x^2  - 1) (x + 5)

 

(x - 0 ) ( x + 1) (x - 1) (x + 5)

 

These are true

 

f(x)  =    when  x  = -5

 

f(x)  ivided by ( x + 5)  as a remainder of 0      [ because ( x + 5) is a factor ]

 

 

cool cool cool

CPhill  Oct 24, 2018
 #2
avatar+90969 
+1

We can make this a little more simple than by testing all possible roots right off

 

One possible rational root is 1

 

If we can add the coefficients and  constant of the polynomial and get 0, then 1 is a root....and this is true

 

So  ( x - 1)  is a factor

 

We can find the remaining polynomial with some division

 

             x^2  + 8x + 15

x - 1  [  x^3 + 7x^2  + 7x  - 15  ]

            x^3  - 1x^2

           ___________________

                      8x^2  + 7x - 15

                      8x^2  - 8x

                     _____________

                               15x  - 15

                               15x -  15

                               ________

 

So the remaining polynomial is

 

x^2 + 8x  + 15

 

The possible rational roots are   ±1  ±3  ±5  ±15

 

However...instead of testing these to see which produce a 0...note that the polynomial can be factored as

 

(x + 3) ( x + 5)

 

So....the factors are   ( x - 1) ( x + 3) ( x + 5)

 

 

 

cool cool cool

CPhill  Oct 24, 2018

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