Question

show that (x+4),(x-3) and (x-7) are factors of x cube 3-6x power of 2-19x+84.

Answer 1

Answer:

Let f(x) = x3 – 6x2 – 19x + 84  If x + 4 = 0, then x = -4  If x – 3 = 0, then x = 3  and if x – 7 = 0, then x = 7 Now,  f(-4) = (-4)3 – 6(-4)2 – 19(-4) + 84  = -64 – 96 + 76 + 84  = 160 – 160  = 0  f(-4) = 0 f(3) = (3)3 – 6(3)2 – 19 x 3 + 84  = 27 – 54 – 57 + 84  = 111 -111 = 0  f(3) = 0  f(7) = (7)3 – 6(7)2 – 19 x 7 + 84  = 343 – 294 – 133 + 84 = 427 – 427  = 0  f(7) = 0  Hence (x + 4), (x – 3), (x – 7) are the factors of f(x).Read

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