Hello, thegersters!
Sorry, you picked the wrong one . . .
Which of the following functions has exactly 4 distinct real zeros?
\(\displaystyle f(x)\:=\:x^6\,-\,1\;\;\;g(x)\:=\:x^3\,-\,x^2\,+\,x\,-\,1\;\;\;h(x)\:=\:x^3\,-\,2x^2\,-\,x\,+\,2\)
\(\displaystyle p(x)\:=\:x^4\,-\,3x^2\,+\,2\;\;\;q(x)\:=\:x^4\,+\,3x^2\,+\,2\;\;\;\text{none of these}\)
We will factor all of them . . .
\(\displaystyle f(x)\:=\:x^6\,-\,1\:=\
x^3\,-\,1)(x^3\,+\,1) \;=\;(x\,-\,1)(x^2\,+\,x\,+\,1)(x\,+\,1)(x^2\,-\,x\,+\,1)\)
. . . Two real zeros:
.\(\displaystyle x\,=\,1,\,-1\)
\(\displaystyle g(x)\:=\:x^3\,-\,x^2\,+\,x\,-\,1\:=\:x^2(x\,-\,1) + (x\,-\,1)\:=\
x\,-\,1)(x^2\,+\,1)\)
. . . One real zeros:
.\(\displaystyle x\,=\,1\)
\(\displaystyle h(x)\:=\:x^3\,-\,2x^2\,-\,x\,+\,2\:=\:x^2(x\,-\,2) - (x\,-\,2)\:=\
x\,-\,2)(x^2\,-\,1)\:=\
x\,-\,2)(x\,-\,1)(x\,+\,1)\)
. . . Three real zeros:
.\(\displaystyle x\,=\,2,\,1,\,-1\)
\(\displaystyle p(x)\:=\:x^4\,-\,3x^2\,+\,2\:=\
x^2\,-\,1)(x^2\,-\,2)\:=\
x\,-\,1)(x\,+\,1)(x\,-\,\sqrt{2})(x\,+\,\sqrt{2})\)
. . . Four real zeros:
.\(\displaystyle x\,=\,1,\,-1,\,\sqrt{2},\,-\sqrt{2}\)
. <-- This one!
\(\displaystyle q(x)\:=\:x^4\,+\,3x^2\,+\,2\:=\
x^2\,+\,1)(x^2\,+\,2)\)
. . . No real zeros.
