Question

show that x+4,.x-3 and x-7 are factors of (x2-6x2-19x+84)​

Answer 1

Hence, 7 is also a root of the given polynomial i.e. (x – 7) will be a factor of it. Hence, it is shown that factors (x + 4), (x – 3), (x – 7) are factors pf x3−6x2−19x+84.

Answer 2

Step-by-step explanation:

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).