Show that (x+4),(x-3) and (x-7) are factors of x3-6x2-19x+84.
Let f (x) = x3-6x2-19x+84 be the given polynomial.
In order to prove that (x + 4), (x – 3) and (x – 7) are factors of f (x), it is sufficient to prove that f (-4) = 0, f (3) = 0 and f (7) = 0 respectively.
Now,
f (x) = x3-6x2-19x+84
f (-4) = (-4)3 – 6 (-4)2 – 19 (-4) + 84
= -64 – 96 + 76 + 84
= 0
f (3) = (3)3 – 6 (3)2 – 19 (3) + 84
= 27 – 54 – 57 + 84
= 0
f (7) = (7)3 – 6 (7)2 – 19 (7) + 84
= 343 – 294 – 133 + 84
= 0
Hence, (x – 4), (x – 3) and (x -7) are the factors of the given polynomial x3-6x2-19x+84.