Use the Rational Root Theorem to list all possible rational roots of the polynomial equation x3 – x2 – x – 3 = 0. Do not find the actual roots. (1 point)
–3, –1, 1, 3
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you're right, but what matters is the process you used.
ok can u check another one for me
What does Descartes's Rule of Signs tell you about the real roots of the polynomial?
–2x3 + 3x2 – 5x – 2 = 0 (1 point)
There is one positive root and either 2 or 0 negative roots.
There are either 2 or 0 positive roots and there are either 2 or 0 negative roots.
There is one positive root and one negative root.
There are either 2 or 0 positive roots and one negative root.
i think it c
@wio am a right
Well it has 3 roots. We know it has 2 or 0 positive roots.
so its d
you didn't list d
aThere is one positive root and either 2 or 0 negative roots.
bThere are either 2 or 0 positive roots and there are either 2 or 0 negative roots.
cThere is one positive root and one negative root.
dThere are either 2 or 0 positive roots and one negative root.
d looks best.
ok i have 1 more
Find a third-degree polynomial equation with rational coefficients that has roots –4 and 6 + i. (1 point)
x3 – 8x2 – 11x + 148 = 0
x3 – 8x2 – 12x + 37 = 0
x3 – 12x2 + 37x = 0
x3 – 8x2 – 11x = 0
i think this one is b or a
Obviously it is a or b, since 0 is not a root.
so which one do u think it is im going with a
can't you just plug 4 into one of them to find out?