Show that (x+4), (x-3) and (x-7) are factors of x³-6x²-19x+84.
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hii friend!!!☺️
x+4=0
x=-4
p(x)=(-4)^3-6(-4)^2-19(-4)+84=0
= -64-96+76+84=0
= 0=0
×-3=0
×=3
p(x)=(3)^3-6(3)^2-19(3)+84=0
=27-54-57+84=0
=0=0
×-7=0
×=7
p(x)=(7)^3-6(7)^2-19(7)+84=0
=343-294-133+84=0
=0=0
therefore(x+4) (x-3)(x-7) are factors of p(x)
hope this will help you and please mark my answer as brainliest
thank u so much!!☺️
x+4=0
x=-4
p(x)=(-4)^3-6(-4)^2-19(-4)+84=0
= -64-96+76+84=0
= 0=0
×-3=0
×=3
p(x)=(3)^3-6(3)^2-19(3)+84=0
=27-54-57+84=0
=0=0
×-7=0
×=7
p(x)=(7)^3-6(7)^2-19(7)+84=0
=343-294-133+84=0
=0=0
therefore(x+4) (x-3)(x-7) are factors of p(x)
hope this will help you and please mark my answer as brainliest
thank u so much!!☺️
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