Question

check whether (x-2),(x+3) and (x-7) are the factors of x³-6x²-13x+42

Answer 1

given :-

p(x) = x³ - 6x² - 13x + 42

◻ for (x - 2)

g(x) = x - 2

➡ x = 2

on putting values,

➡ p(2) = (2)³ - 6(2)² - 13(2) + 42

= 8 - 24 - 26 + 42

= -42 + 42

= 0

since the result came 0, i.e (x - 2) is a factor of p(x)

◻ for (x + 3)

g(x) = x + 3

➡ x = -3

on putting values,

➡ p(-3) = (-3)³ - 6(-3)² - 13(-3) + 42

= -27 - 54 + 39 + 42

= -81 + 81

= 0

result is 0. hence (x + 3) is also a factor of p(x)

◻ for (x - 7),

g(x) = x - 7

➡ x = 7

on putting values,

➡ p(7) = (7)³ - 6(7)² - 13(7) + 42

= 343 - 294 - 91 + 42

= -42 + 42

= 0 $<p> hence, (x + 7) is a factor of p(x). </p>$

Answer 2

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