Sabe-se que -1 é raiz dupla do polinômio P(x) = 2x4 + x3 - 3x2 - x + 1. As outras raízes são númerosa)imaginários puros. b) reais negativos.c)

\[2x^4+ x^3 - 3x^2 - x + 1\,\,| \underline{\,\,x^2+2x+1}\]

\[\eqalign{2x^4+ x^3 - 3x^2 - x + 1\,\,| &\underline{\,\,x^2+2x+1} \\ &\color{Blue}{2x^2} }\]

Multiplicando \(2x^2\) por \(x^2+2x+1\) e realizando a subtração, tem-se o seguinte:

\[\eqalign{&\,\,\,\,\,\,\,2x^4+ x^3 - 3x^2 - x + 1\,\,| \underline{\,\,x^2+2x+1} \\ &\underline{\color{Blue}{-(2x^4+4x^3+2x^2)} }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{Black}2x^2 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{Blue}{-3x^3-5x^2-x+1} }\]

\[\eqalign{&\,\,\,\,\,\,\,\,2x^4+ x^3 - 3x^2 - x + 1\,\,| \underline{\,\,x^2+2x+1} \\ &\underline{-(2x^4+4x^3+2x^2)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x^2\color{Blue}{-3x} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3x^3-5x^2-x+1 }\]

Multiplicando \(-3x\) por \(x^2+2x+1\) e realizando a subtração, tem-se o seguinte:

\[\eqalign{&\,\,\,\,\,\,\,\,2x^4+ x^3 - 3x^2 - x + 1\,\,| \underline{\,\,x^2+2x+1} \\ &\underline{-(2x^4+4x^3+2x^2)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x^2-3x \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3x^3-5x^2-x+1 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\color{Blue}-(-3x^3-6x^2-3x)} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{Blue}{x^2+2x+1} }\]

\[\eqalign{&\,\,\,\,\,\,\,\,2x^4+ x^3 - 3x^2 - x + 1\,\,| \underline{\,\,x^2+2x+1} \\ &\underline{-(2x^4+4x^3+2x^2)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x^2-3x\color{Blue}{+1} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3x^3-5x^2-x+1 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-(-3x^3-6x^2-3x)} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x^2+2x+1 }\]

Multiplicando \(x^2\) por \(x^2+2x+1\) e realizando a subtração, tem-se o seguinte:

\[\eqalign{&\,\,\,\,\,\,\,\,2x^4+ x^3 - 3x^2 - x + 1\,\,| \underline{\,\,x^2+2x+1} \\ &\underline{-(2x^4+4x^3+2x^2)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x^2-3x+1 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3x^3-5x^2-x+1 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-(-3x^3-6x^2-3x)} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x^2+2x+1 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\color{Blue}{-(x^2+2x+1)}} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \color{Blue}{0} }\]

Com quociente \(Q=2x^2-3x+1\) e resto \(R=0\), o resultado da divisão de \(P(x)=2x^4+ x^3 - 3x^2 - x + 1\) por \((x+1)^2= x^2 + 2x + 1\) é:

\[\eqalign{ {2x^4+ x^3 - 3x^2 - x + 1 \over x^2+2x+1} &= Q+R \\ &= 2x^2-3x+1+0 \\ }\]

Portanto, \(P(x)\) fica da seguinte forma:

\[\eqalign{ 2x^4+ x^3 - 3x^2 - x + 1&=(x^2+2x+1)(2x^2-3x+1) \cr &=(x+1)^2(2x^2-3x+1) }\]

As outras raízes referentes no enunciado são as raízes do quociente \(2x^2-3x+1\), que está no formato \(ax^2+bx+c\). Portanto, com \(a=2\), \(b=-3\) e \(c=1\), as outras raízes são calculadas da seguinte forma:

\[x={-b \pm \sqrt{b^2-4ac} \over 2a}\]

Portanto, pelo Método de Bhaskara, as raízes de \(2x^2-3x+1\) são:

\[\eqalign{ x&={-(-3) \pm \sqrt{(-3)^2-4\cdot 2\cdot 1} \over 2\cdot 2} \\ &={3 \pm \sqrt{9-8} \over 4} \\ &={3 \pm 1 \over 4} \to \left\{ \begin{matrix} x_1=1 \\ x_2={1 \over 2} \end{matrix} \right. }\]

Ou seja, além da raiz dupla igual a \(-1\), \(P(x)\) também possui raízes igual a \(1\) e \({1 \over 2}\), ambas raízes racionais.

Resposta correta: d) racionais.