Show that (x+4), (x-3) and (x-7) are factors of ​.

Hello!

x³ - 6x² - 19x + 84 =

= x³ - 3x² - 3x² + 9x - 28x + 84 =

= x² × (x - 3) - 3x × (x - 3) - 28 × (x - 3) =

= (x - 3) × (x² - 3x - 28) =

= (x - 3) × (x² + 4x - 7x - 28) =

= (x - 3) × [ x × (x + 4) - 7 × (x + 4)] =

= (x - 3) × (x + 4) × (x - 7)

Good luck! :)

Step-by-step explanation:

Let p(x)= x³-6x²-19x+84 be the polynomial

Let us take these factors (x+4), (x-3) and (x-7) =0

Therefore,

x=-4, 3 ,7

Substitute it with p(x)

p(-4)= (-4)³-6(-4)²-19(-4)+84

=-64-96+76+84

=-160+160

=0

Therefore ,(x+4) is the factor of p(x)

p(3)=(3)³-6(3)²-19(3)+84

=27-54-57+84

=0

Therefore ,(x-3) is a factor of p(x)

p(7)=(7)³-6(7)²-19(7)+84

=343-294-133+84

=0

Therefore, (x-7) is a factor of p(x)

We proved that (x+4), (x-3) and (x-7) are factors of x³-6x²-19x+84

you would have to sell 24 bracelets.

step-by-step explanation:

84/3.5 = 24