Question

Find the zeros of cubic polynomial 3x^3-2x^2-19x-6

Answer 1

3x³-2x²-19x-6
Factors of 6=±1,±2,±3&±6
Let p(x)=1
:.3(1)³-2(1)²-19(1)-6=3-2-19-6=-24≠0
Let p(x)=-1
:.3(-1)³-2(-1)²-19(-1)-6=-3-2+19-6=8≠0
Let p(x)=2
:.3(2)³-2(2)²-19(2)-6=3×8-2×4-19×2-6=24-8-38-6=-28≠0
Let p(x)=-2
:.3(-2)³-2(-2)²-19(-2)-6=3×-8-2×4+38-6=-24-8+38-6=0
:.(x+2) is factor
Now dividing 3x³-2x²-19x-6 by (x+2)
:. Answer=(x+2)(x-3)(3x+1)

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