Show that $(x - 1)$ is a factor of $x^{3} - 7 x^{2} + 14 x - 8$ . Hence, completely factorise the given expression.

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Solution

Let $f (x) = x^{3} - 7 x^{2} + 14 x - 8$
Then, for $x = 1$
$f (1) = (1)^{3} - 7 (1)^{2} + 14 (1) - 8 = 1 - 7 + 14 - 8 = 0$
Thus, $(x - 1)$ is a factor of $f (x) .$
Now, performing long division we have
Hence, $f (x) = (x - 1) (x^{2} - 6 x + 8)$
$= (x - 1) (x^{2} - 4 x - 2 x + 8)$
$= (x - 1) [x (x - 4) - 2 (x - 4)]$
$= (x - 1) (x - 4) (x - 2)$

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Show that (x–1) is a factor of x3–7x2+14x–8. Hence, completely factorise the given expression.

Let f(x)=x3–7x2+14x–8Then, for x=1f(1)=(1)3–7(1)2+14(1)–8=1–7+14–8=0Thus, (x–1) is a factor of f(x).Now, performing long division we haveHence, f(x)=(x–1)(x2–6x+8)=(x–1)(x2–4x–2x+8)=(x–1)[x(x–4)−2(x–4)]=(x–1)(x–4)(x–2)

Show that $(x - 1)$ is a factor of $x^{3} - 7 x^{2} + 14 x - 8$ . Hence, completely factorise the given expression.