Skoro wielomian ma dwa pierwiastki całkowite tego samego znaku to znaczy, że tymi pierwiastkami są liczby 1 i 2 lub -1 i -2.

$\{\begin{array}{l}w\left(1\right)=0\\ w\left(2\right)=0\end{array}$   

$\{\begin{array}{l}1+a-5+b+2=0\\ 16+8a-20+2b+2=0\end{array}$  

$\{\begin{array}{l}a+b=2\\ 8a+2b=2\mid :2\end{array}$  

$\{\begin{array}{l}b=2-a\\ 4a+2-a=1\end{array}$  

$\{\begin{array}{l}b=2-a\\ 3a=-1\end{array}$  

$\{\begin{array}{l}b=2-a\\ a=-\frac{1}{3}\end{array}$  

$\{\begin{array}{l}b=\frac{7}{3}\\ a=-\frac{1}{3}\end{array}$  

lub

$\{\begin{array}{l}w\left(-1\right)=0\\ w\left(-2\right)=0\end{array}$  

$\{\begin{array}{l}1-a-5-b+2=0\\ 16-8a-20-2b+2=0\end{array}$  

$\{\begin{array}{l}-a-b=2\\ -8a-2b=2\mid :\left(-2\right)\end{array}$  

$\{\begin{array}{l}a+b=-2\\ 4a+b=-1\end{array}$  

$\{\begin{array}{l}b=-a-2\\ 4a-a-2=-1\end{array}$  

$\{\begin{array}{l}b=-a-2\\ 3a=1\end{array}$  

$\{\begin{array}{l}b=-\frac{7}{3}\\ a=\frac{1}{3}\end{array}$  

Odpowiedź B, D.