Show that (x - 1) is a factor of x3 – 7x 2 + 14x – 8. Hence completely factorize
(ICSE 2007)
Answer:
Let: F(x)
= x3 –
7x 2 + 14x – 8
Since (x
- 1) is a factor of F(x), therefore, according to Remainder theorem,
we have,
F (1) = 0
Now, F (1)
= (1)3 - 7(1)2 + 14(1) – 8
Or, F(1) =1
– 7 +14 - 8 = 0
So, (x – 1) is a factor.
Now we will divide (x3 – 7x 2 + 14x – 8)/(x
– 1)
(x - 1) // x3 –
7x 2 + 14x – 8 // x 2 -
6x + 8
x3
- x 2
- 6 x 2 + 14x
- 6 x
2 + 6x
8x – 8
8x – 8
0 0
(x 2 -
6x +8) is the other factor of F(x). We will further simplify, thus:
x 2 -
6x +8 = x 2 - 4x - 2x + 8 = x(x – 4) – 2(x – 4) = (x – 2)(x –
4)
So, x3 –
7x 2 + 14x – 8 = (x – 1 )(x
– 2)(x – 4)
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