Question

Two zeroes of the polynomial ax^3+3x^2-bx-6 are -1 and -2,find the vues of a and b also find the third zero

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

p(x)=ax^3+3x^2-bx-6

p(-1)=a(-1)^3+3(-1)^2-b(-1)-6=0

-a+3+b-6=0

-a+b=3---(1)

p(-2)=a(-2)^3+3(-2)^2-b(-2)-6=0

-8a+12+2b-6=0

-8a+2b=-6

-4a+b=-3--(2)

solve (1) and (2)

3a=6

a=2

b=5

the polynomial is 2x^3+3x^2-5x-6

(x+1)(x+2)(2x-3)

the other zero is 3/2