+128474 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 ]
+128474 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
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15x - 15
15x - 15
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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)
