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Use the given factor and division to solve the equation. show all work x^3+4x^2+x-6=0 ; (x-1)

User Barbarian
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\begin{gathered} x^3+4x^2+x-6=(x-1)(ax^2+bx+c) \\ \\ x^3+4x^2+x-6=ax^3+bx^2+cx-ax^2-bx-c \\ x^3+4x^2+x-6=ax^3+x^2(b-a)+x(c-b)-c \\ a=1 \\ b-a=4 \\ c-b=1 \\ -c=-6 \\ \\ a=1 \\ b=5 \\ c=6 \\ \\ \\ x^3+4x^2+x-6=(x-1)(x^2+5x+6) \\ =(x-1)(x+3)(x+2) \\ \\ \text{ }x^3+4x^2+x-6=0\text{ } \\ (x-1)(x+3)(x+2)=0 \\ \\ \text{thus } \\ x=1\text{ , x=-3 or x=-2 } \\ \\ 1,-3,-2\text{ are the solutions of the equation } \end{gathered}

User Perel
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