Verified answer

Ответ:

[tex]x^4+3x^3-3x^2-7x+6=(x+3)(x+2)(x-1)(x-1)[/tex]

Объяснение:

[tex]W(x)=x^4+3x^3-3x^2-7x+6[/tex]

[tex]W(1)=0[/tex]

[tex]W(x)=Q_1(x)\cdot P_1(x)+R_1[/tex]

[tex]P_1(x)=x-1[/tex]

[tex]{\begin{array}{|r|r|r|r|r|r|}\cline{1-6}&a_4&a_3&a_2&a_1&a_0 \\ &1&3&-3&-7&6 \\ \cline{1-6} 1&&1&4&1&-6 \\ \cline{1-6} &1&4&1&-6&0 \\ &b_3&b_2&b_1&b_0&R_1\\ \cline{1-6} \end{array}}[/tex]

[tex]W(x)=x^{4} + 3 x^{3} - 3x^{2} - 7 x + 6=(x^{3} + 4x^{2} + x - 6) (x - 1)[/tex]

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[tex]Q_1=x^{3} + 4x^{2} + x - 6[/tex]

[tex]Q_1(1)=0[/tex]

[tex]Q_1(x)=Q_2(x)\cdot P_2(x)+R_2[/tex]

[tex]P_2(x)=x-1[/tex]

[tex]{\begin{array}{|r|r|r|r|r|}\cline{1-5}&a_3&a_2&a_1&a_0 \\ &1&4&1&-6 \\ \cline{1-5} 1&&1&5&6 \\ \cline{1-5} &1&5&6&0 \\ &b_2&b_1&b_0&R_2\\ \cline{1-5} \end{array}}[/tex]

[tex]Q_1(x)=x^{3} + 4x^{2} + x - 6=(x^{2} + 5 x + 6)(x - 1)[/tex]

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[tex]Q_2=x^2 + 5 x + 6[/tex]

[tex]Q_2(-2)=0[/tex]

[tex]Q_2(x)=Q_3(x)\cdot P_3(x)+R_3[/tex]

[tex]P_3(x)=x+2[/tex]

[tex]{\begin{array}{|r|r|r|r|}\cline{1-4} &a_2&a_1&a_0 \\ &1&5&6\\ \cline{1-4} -2&&-2&-6 \\ \cline{1-4} &1&3&0 \\&b_1&b_0&R_3\\ \cline{1-4} \end{array}}[/tex]

[tex]Q_2(x)=x^{3} + 4x^{2} + x - 6=(x+3)(x+2)[/tex]

[tex]x^4+3x^3-3x^2-7x+6=(x^{3} + 4x^{2} + x - 6) (x - 1)=(x^{2} + 5 x + 6)(x - 1)(x-1)=(x+3)(x+2)(x-1)(x-1)[/tex]