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[tex] \: \huge\mathfrak\red{answer} \: [/tex]
= (x – 1)(x + 1)(x – 5).
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Let x – 1 = 0, then x = 1
Substituting the value of x in f(x),
f(x) = x3 – 5x2 – x + 5
= (1)3 – 5(1)2 – 1 + 5
= 1 – 5 – 1 + 5
= 0
∵ Reminder = 0
∴ x – 1 is a factor of x3 – 5x2 – x + 5
Now dividing f(x) by x – 1, we get
(x-1)x3-5x2-x+5¯(x2-4x-5
x3 – x2
– +
______________________
-4x2 – x
– 4x2 + 4x
+ –
_______________________
– 5x + 5
–5x + 5
+ – x
∴ x3 – 5x2 – x + 5
= (x – 1)(x2 – 4x – 5)
= (x – 1)[x2 – 5x + x – 5]
= (x – 1)[x(x – 5) + 1(x – 5)]
= (x – 1)(x + 1)(x – 5).
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