Question

Show that (x – 1) is a factor of x^3 − 7x^2 + 14x – 8

Answer 1

SOLUTION :

Let

$\bf P(x) = x^{3}-7x^{2}+14x-8$

and

$\bf g(x) = x-1$

⇒ x = 1

Now,

$\bf P(1) = (1)^{3}-7(1)^{2}+14(1)-8$

       

      $\bf= 1-7+14-8$

      $\bf= 1+14-8-7$

      $\bf= 15-15$

      $\bf= 0$

Hence

g(x) is the factor of P(x)

Answer 2

$<b>$Here is the answer to your question!!

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Let p (x) = x³ - 7x² + 14x - 8

and f(x) = x - 1
= x - 1


Now,

p(1) = (1)³ -7(1)² + 14 (1) - 8

= 1 - 7 + 14 - 8

= 1 + 14 -8 -7

= 15 -15 = 0


Hence,

f(x) is the factor of p(x)


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Hope it helped!!