anonymous
  • anonymous
Alg 2 help Use the Rational Root Theorem to list all possible rational roots of 3x3 + x2 – 15x – 5 = 0. Then find any actual rational roots. Find the roots of 10x4 + x3 + 7x2 + x – 3 = 0 A polynomial equation with rational coefficients has the roots Sq rt of 7 and –5i. Find two additional roots Find a third-degree polynomial equation with integer coefficients that has the roots 8 and 3i.
Algebra
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The rational root theorem states that any rational root of a polynomial will be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. The leading coefficient is 10, so the factors are -10, -5, -2, -1, 1, 2, 5, and 10. The constant term is -3, so the factors are -3, -1, 1, and 3. The possible rational roots are -3/10, -1/10, -3/5, -1/5, -3/2, -1/2, -3/1, -1/1, 1/1, 3/1, 1/2, 3/2, 1/5, 3/5, 1/10, and 3/10.
anonymous
  • anonymous
can you be more clear i still dont know the answer
anonymous
  • anonymous
I listed a big list of fractions at the end, those are all the possible roots from the rational root theorem. (I know, it's a big list)

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anonymous
  • anonymous
Oh oops I found the roots of the wrong equation
anonymous
  • anonymous
The leading coefficient is 3 and the constant term is -5, so the roots are -5/3, -5/1, -1/1, 1/1, 5/1, and 5/3
anonymous
  • anonymous
which number is this, one right?
anonymous
  • anonymous
@sauravshakya do you understand this??
precal
  • precal
you should post your questions one at a time because it will be confusing for you to understand our postings
anonymous
  • anonymous
what unit and lesson?
anonymous
  • anonymous
Theorems About Roots of Polynomial Equations
anonymous
  • anonymous
1. 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) (1 pt) –3, –1, 1, 3 (0 pts) 1, 3 (0 pts) –33 (0 pts) no roots 1 /1 point This item has been reviewed and is scheduled to be updated. All students will receive full credit for any response to the following. 2. Find the roots of the polynomial equation. 2x3 + 2x2 – 19x + 20 = 0 (1 point) (1 pt) , , –4 (1 pt) , , 4 (1 pt) , , –4 (1 pt) , , 4 1 /1 point 3. A cubic polynomial with rational coefficients has the roots 6 +and . Find one additional root. (1 point) (0 pts) –6 – (0 pts) 6 + (0 pts) –6 + (1 pt) 6 – 0 /1 point 4. Find a third-degree polynomial equation with rational coefficients that has roots –4 and 6 + i. (1 point) (1 pt) x3 – 8x2 – 11x + 148 = 0 (0 pts) x3 – 8x2 – 12x + 37 = 0 (0 pts) x3 – 12x2 + 37x = 0 (0 pts) x3 – 8x2 – 11x = 0 1 /1 point 5. What does Descartes's Rule of Signs tell you about the real roots of the polynomial? –2x3 + 3x2 – 5x – 2 = 0 (1 point) (0 pts) There is one positive root and either 2 or 0 negative roots. (0 pts) There are either 2 or 0 positive roots and there are either 2 or 0 negative roots. (0 pts) There is one positive root and one negative root. (1 pt) There are either 2 or 0 positive roots and one negative root. 0 /1 point
anonymous
  • anonymous
no i think its unit 6 lesson 5 it only has four questions
anonymous
  • anonymous
1. What is the solution of the equation? –9 = –7 (1 point) (1 pt) –6 (0 pts) 4 (0 pts) 14 (0 pts) –8 1 /1 point 2. What is the solution of the equation? – 5 = 59 (1 point) (1 pt) –5, 11 (0 pts) 5 (0 pts) 11 (0 pts) –11 1 /1 point 3. The area of a circular trampoline is 108.94 square feet. What is the radius of the trampoline? Round to the nearest hundredth. (1 point) (0 pts) 10.44 feet (1 pt) 5.89 feet (0 pts) 34.68 feet (0 pts) 3.32 feet 1 /1 point 4. What is the solution of the equation? 5x = (1 point) (0 pts) –1 (0 pts) 1 and (1 pt) 1 (0 pts) 1 /1 point 5. What is the solution of – = 5 ? (1 point) (0 pts) x = 0 (0 pts) x = 16 and x = 0 (1 pt) x = 16 (0 pts) x = 16 and x = 1 1 /1 point
anonymous
  • anonymous
lol its the same questions as the questions listed
anonymous
  • anonymous
hhmmm idk, i cant find em
anonymous
  • anonymous
its ok thank you

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