Question

Show that (x+4)(x-3) and (x-7) are factors of x^3-6x^2 19x+84​

Answer 1

Answer:

p(x) = x³-6x²-19x+84

(i) g(x) = (x+4)

   g(x) = 0

   x+4 = 0

    x = -4

p(-4) = (-4)³- 6×(-4)²- 19×(-4) + 84

        = -64 - 6×16 - (-76) +84

        = -64 - 96 + 76 +84

        = -160 + 160

        = 0

∴ (x+4) is a factor

(ii) g(x) = (x-3)

    g(x) = 0

    x-3 = 0

    x = 3

p(3) = (3)³ - 6×(3)²- 19×(3) + 84

      = 27 - 54 - 57 + 84

      = 111 - 111

      = 0

∴ (x-3) is a factor

(iii) g(x) = (x-7)

     g(x) = 0

     x-7 = 0

     x = 7

p(7) = (7)³- 6×(7)²- 19(7) +84

      = 343 - 294 - 133 +84

      = 427 - 427

      = 0

∴ (x-7) is a factor

Hope this helps!!!!!