What can we afirm about the roots of 2x³ - x² - 2x + 1 = 0 ? a) only one of them is irrational b) they belong to [-1,1] c) two of them are negative d) none of them is real Can someone help me? I divided it by x - 1 and got x = -1 and x = 1/2. Is it right to say they belong to [-1,1]? Because option 'a' also seems right. Help!
2x³ - x² - 2x + 1 = 0 has 3 roots x=1 x=-1 x=1/2
Still, both a and b seem right.
When x=-1 y=0 , it cant be b)
a
Yep, it can be b as well. why do you think its a?
When x=-1 y=0 , it cant be b
For [-1,1]
I know that. Why did you post 'a'?
b) they belong to [-1,1] No roots for this polynomial are irrational. Do you know the Rational Roots Theorem, Synthetic Division? I don't know the divisor you used for your division or the dividend but if you provide that information or perhaps use the "Draw" button to write your division, I will try to help you.
After finding 2x² + x - 1 I did this: x = - 1 + sqrt9 / 4 x = -1 - sqrt9 /4 Then I got x = -1 and x = 1/2
1/2 is irrational, isn't it?
oops haha
1/2 is rational. A rational number is one that can be written as the ratio of 2 integers. The first 5 letters of ratio nal are a reminder.
Yep, sorry.
What did you divide by to get 2x² + x - 1? I'm trying to tune in to your thinking. I just saw this thread.
I divided 2x³ - x² - 2x + 1 by x - 1
remainder's 0 so I guess it's right
Okay, I got the same quotient. 2x^2 + x -1 should factor.
(2x - 1) (x + 1) = 0 2x - 1 = 0 or x + 1 =0 x = 1/2 or x = -1
Thanks. I can see it clearly now, it's b indeed.
You understand the [-1,1] notation. Yes?
Yep. Thx a lot
Great.
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