Question

show that (x+4),(x+3) and (x-7) are the factors of x^3-6x^2-19x+84 PLEASE HELP ME NOW FRIENDS PLS☹️​

Answer 1

Answer:

f(x)=x^3-6x^2-19x+84.

f(-4)=(-4)^3-6(-4)^2-19(-4)+84=-64-96+76+84=0

so (x+4)is a factor of f(x).

similarly if we may found f(-3)= 0 and f(7)=0 then we say that (x+3) and (x-7)is factor of f(x).

Answer 2

Step-by-step explanation:

Here,

in first equation is (x+4)

that is. x=-4

so, putting value of x=-4 in given polynomial

:. (-4)³-6(-4)²-19(-4)+84. =0

= -64-96+76+84=0

= -160+160=0

yes it is a factor of given polynomial

in second equation is (x+3)

that is x=-3

putting value of x=-3 in given polynomial

=(-3)³-6(-3)²-19(-3)+84

=-27-54+57+84

= -81+141 is not equal to 0

it is not a factor of given polynomial

in third equation is (x-7)

that is x=7

putting value of x=7 in given polynomial

=(7)³-6(7)²-19(7)+84

=343-294-133+84

=427-427=0

yes it is a factor of given polynomial

I hope this answer is correct