Question

show that [x+4],[x-3]and [x-7] are the factors of x^3-6x^2-19x+84\​

Answer 1

Answer:

$(x+4),\ (x-3),\ (x-7),$  are all factors of $p(x)=x^{3} -6x^{2} -19x+84$.

Step-by-step explanation:

Given $p(x)=x^{3} -6x^{2} -19x+84$.

$x+4$ is a factor of the polynomial if $p(x)=0$ for $x=-4$  (that is $x+4=0$ implies $x=-4$)

Therefore $p(-4)=(-4)^{3} -6(-4)^{2} -19(-4)+84$

                           $=-64-96+76+84=-160+160=0$

Hence $x+4$ is a factor .

similarly $x-3$ is a factor if $p(3)=0$.

Therefore $p(3)=3^{3} -6(3^{2} )-19(3)+84$

                        $=27-54-57+84=111-111=0$

Hence $x-3$ is a factor.

also if $x-7$ is a factor, then $p(7)=0$

$p(7)$

       $=7^{3} -6(7^{2} )-19(7)+84\\\\=343-294-133+84\\\\=427-427=0$

Hence $x-7$ is also a factor of the given polynomial

Hence the answer

thank you