Odpowiedź:
[tex]W(x)=4(x-3)^2(x+\frac{1}{2})(x-\frac{1}{2})[/tex]
Szczegółowe wyjaśnienie:
[tex]W(x)=(2x^2-5x-3)(2x^2-7x+3)[/tex]
Najpierw zajmiemy się "pierwszym nawiasem":
[tex]\Delta=(-5)^2-4\cdot 2\cdot (-3)=25+24=49\\\\\sqrt{\Delta}=\sqrt{49}=7[/tex]
[tex]x_1=\frac{-(-5)+7}{2\cdot 2}=\frac{5+7}{4}=\frac{12}{4}=3\\\\x_2=\frac{-(-5)-7}{2\cdot 2}=\frac{5-7}{4}=\frac{-2}{4}=-\frac{1}{2}[/tex]
[tex]2x^2-5x-3=2(x-3)(x-(-\frac{1}{2}))=2(x-3)(x+\frac{1}{2})[/tex]
Teraz "drugi nawias":
[tex]\Delta=(-7)^2-4\cdot 2\cdot 3=49-24=25\\\\\sqrt{\Delta}=\sqrt{25}=5[/tex]
[tex]x_1=\frac{-(-7)+5}{2\cdot 2}=\frac{7+5}{4}=\frac{12}{4}=3\\\\x_2=\frac{-(-7)-5}{2\cdot 2}=\frac{7-5}{4}=\frac{2}{4}=\frac{1}{2}[/tex]
[tex]2x^2-7x+3=2(x-3)(x-\frac{1}{2})[/tex]
Ostatecznie:
[tex]W(x)=2(x-3)(x+\frac{1}{2})\cdot 2(x-3)(x-\frac{1}{2})=4(x-3)^2(x+\frac{1}{2})(x-\frac{1}{2})[/tex]
