Show that (x+4), (x-3) and (x-7)
factors of x² - 6x0² - 19x +84
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if (x+4), (x-3) and (x-7) are the factors of p(x)= x3 - 6x^2- 19x +84 then p(-4),p(3),p(7) equal to zero.
p(-4)= ( -4)^3 - 6(-4)^2 -19(-4) +84 = 0
similarly, p(3)=0 and p(7)=0
Therefore, (x+4), (x-3) and (x-7) are the factors of p(x)= x3 - 6x^2- 19x +84
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