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Question
Show that (x – 3) is a factor of x3 – 7x2 + 15x – 9. Hence factorise x3 – 7x2 + 15 x – 9
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Solution
Let x – 3 = 0, then x = 3,
Substituting the value of x in f(x),
f(x = x3 - 7x2 + 15x – 9
= (3)3 – 7(3)2 + 15(3) – 9
= 27 – 63 + 45 – 9
= 72 – 72
= 0
∵ Remainder = 0
∴ x – 3 is a factor of x3 – 7x2 + 15x – 9
Now dividing it by x – 3, we get
`x - 3")"overline(x^3 - 7x^2 + 15x – 9)("x^2 - 4x + 3`
x3 – 3x2
– +
– 4x2 – 15x
– 4x2 + 12x
+ –
3x – 9
3x – 9
– +
x _
∴ x3 – 7x2 + 15x – 9
= (x – 3)(x2 – 4x + 3)
= (x – 3)[x2 – x – 3x + 3]
= (x – 3)[x(x – 1) – 3(x – 1)]
= (x – 3)(x – 1)(x – 3)
= (x – 3)2 (x – 1).
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