## anonymous 4 years ago 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 1, 3 –33 no roots

1. anonymous

i think it is a am i right yes or no

2. anonymous

$a_n = 1 \quad a_0 = -3$

3. anonymous

$\pm \frac{1,3}{1} = 1, 3, -1, -3$

4. anonymous

you're right, but what matters is the process you used.

5. anonymous

ok can u check another one for me

6. anonymous

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

7. anonymous

@wio am a right

8. anonymous

Well it has 3 roots. We know it has 2 or 0 positive roots.

9. anonymous

so its d

10. anonymous

you didn't list d

11. anonymous

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.

12. anonymous

d looks best.

13. anonymous

ok i have 1 more

14. anonymous

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

15. anonymous

Obviously it is a or b, since 0 is not a root.

16. anonymous

so which one do u think it is im going with a

17. anonymous

can't you just plug 4 into one of them to find out?

18. anonymous

whats number 2 to this ?

19. anonymous

a a d a d