distinct real zeros

[thegersters]

Which of the following functions has exactly 4 distinct real zeros?

f(x) = x^6 - 1
g(x) = x^3 - x^2 + x - 1
h(x) = x^3 - 2x^2 - x + 2 Would this be it????
p(x) = x^4 - 3x^2 + 2
p(x) = x^4 + 3x^2 + 2
none of these

 

[soroban]

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.