Question

show that (x-2), (x+3) and (x+7) are factors of x³ - 6x² - 13x + 42

Answer 1

I think there is some error in your question

→ (x-7) is a factor of given p(x)


→3rd Case:

$p(7) = {7}^{3 } - 6(7)^{2} - 13(7) + 42 \\ \\ = 343 - 294 - 91 + 42 = 0\\ \\ = 49 + 42 - 91 = 0 \\ \\ = 91 - 91 = 0$
→ for 1st and 2nd Case refer to attachment

→ we can see that in 1st, 2nd and 3rd Case,

LHS=RHS

Hence verified//

____________________

Answer 2

Hey there!

We know,
If (x-a) is a factor of f(x) then f(a) = 0

So,
f(2) = 2³ - 6(2)² - 13(2) + 42
      = 8 - 24 - 26 + 42 
      = 50 - 50 
      = 0
f(-3) = (-3)³ - 6(-3)² - 13(-3) + 42
        = -27 - 54 + 39 + 42 
        =81 - 81 
        =0
f(-7) = (-7)³ -7(-7)² - 13(-7) + 42
        = -343 - 343 + 91 + 42
        = -686 + 133
        = -553

So, (x-2), (x+3) and (x+7) are not the factors of x³ - 6x² - 13x + 42.

But, if t the question is Show that (x-2), (x+3) and (x-7) are factors of x³ - 6x² - 13x + 42.. Then, (x-2), (x+3) and (x-7) are factors of x³ - 6x² - 13x + 42.

As,
f(7) = 7³ - 6(7)² - 13(7) + 42
      =343-294-91+42 
      = 385 - 385
      = 0

Like this,  (x-2), (x+3) and (x-7) are factors of x³ - 6x² - 13x + 42.