N = 3; 1 and 5i are zeros;f(-l) = -104 n = 3; 4 and 2i are zeros;f(-l) = -50 n =

Question

N = 3; 1 and 5i are zeros;f(-l) = -104 n = 3; 4 and 2i are zeros;f(-l) = -50 n = 3; -5 and 4 + 3i are zeros; f(2) = 91 n = 3; 6 and -5 + 2i are zeros; f(2) = -636 n = 4; i and 3i are zeros;f(-1) = 20 n = 4; -2, - 1/2, and i are zeros;f(-1) = 18 n = 4; -2, 5, and 3 + 2i are zeros;f(1) = -96 n = 4; -4, 1/3, and 2 + 3i are zeros; f(1) = 100 use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x) = x^3 + 2x^2 + 5x + 4 f(x) = x^3 + 7x^2 + x + 7 f(x) = 5x^3 - 3x^2 + 3x - 1 f(x) = -2x^3 + x^2 - x + 7 f(x) = 2x^4 - 5x^3 - x^2 - 6x + 4 f(x) = 4x^4 - x^3 + 5x^2 - 2x - 6

Explanation / Answer

(33) given f(x)=x^3+2x^2+5x+4

here are no sign changes, so there are no positive roots

Now look at f(-x)=(-x)^3+2(-x)^2+5(-x)+4=-x^3+2x^2-5x+4

here there are 3 times sign changes so there as many as three negative root possible, that may be repeated or imaginary

so, there are no positive roots, and there are three or one negative roots

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.