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# Algebra 2 help

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Q1)

Q2)

Q3)

thank you!!

Oct 24, 2018

#1
+100571
+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 ]

Oct 24, 2018
#2
+100571
+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)

Oct 24, 2018