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Question
Show that (x – 1) is a factor of x3 – 5x2 – x + 5 Hence factorise x3 – 5x2 – x + 5.
Solution
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")"overline(x^3 - 5x^2 - x + 5)("x^2 - 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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