Question

x cube minus 3x square minus 9x minus 5​

Answer 1

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x^3+-3x^2-9x-5

Factors of 5:- +-1, +-5 (Trial method)

p(x) = x^3-3x^2-9x-5

p(1) = (1)^3-3×1^2-9×1-5

= 1-3-9-5

= -16

Hence, -16 is not equal to 0

Now,

p(-1) = (-1)^3-3×(-1)^2-9×(-1)-5

= -1-3+9-5

= -9+9

= 0

Therefore, -1 is the zeros of p(x).

(x+1) is the factor of p(x).

Now,

when x^3-3x^3-9x-5 is divided by the factor.......(x+1), the quotient will be = x^2-4x-5

Since,

(x+1)(x^2-4x-5) = x^3-3x^2-9x-5

p(x) = (x+1)(x^2-4x-5)

= (x+1)(x^2-5x+x-5)

= (x+1){x(x-5)+1(x-5)}

= (x+1)(x-5)(x+1)

Hope it might helped u

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