D deedie_06 New member Joined Apr 19, 2007 Messages 1 Apr 19, 2007 #1 1.Find the integer-valued upper and lower bounds to the zeros of the polynomial function f(x)=6x^4-7x^3-26x^2+7x+20. 2. Find the polynomial of degree 3 with the real coefficients that satifies the given condition zeros: 3,i,-i; P(2)=0.
1.Find the integer-valued upper and lower bounds to the zeros of the polynomial function f(x)=6x^4-7x^3-26x^2+7x+20. 2. Find the polynomial of degree 3 with the real coefficients that satifies the given condition zeros: 3,i,-i; P(2)=0.
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,550 Apr 19, 2007 #2 What have you tried? Where are you stuck? 1) You've applied the Rational Roots Test and... then what? 2) You've converted from "x = a" (zero) form to "x - a" (factor) form, multiplied the factors, and... then what? Please be complete and specific. Thank you. Eliz.
What have you tried? Where are you stuck? 1) You've applied the Rational Roots Test and... then what? 2) You've converted from "x = a" (zero) form to "x - a" (factor) form, multiplied the factors, and... then what? Please be complete and specific. Thank you. Eliz.
M morson Full Member Joined Apr 12, 2007 Messages 263 Apr 20, 2007 #3 For the second question, I do not think that a polynomial satisfies those conditions. Or I'm just too sleepy. You have told us that there are 4 zeros to a cubic (and by the fundamental theorem of algebra, we know there are only 3 complex zeros).
For the second question, I do not think that a polynomial satisfies those conditions. Or I'm just too sleepy. You have told us that there are 4 zeros to a cubic (and by the fundamental theorem of algebra, we know there are only 3 complex zeros).
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,550 Apr 20, 2007 #4 morson said: You have told us that there are 4 zeros to a cubic... Click to expand... Good catch! I only noticed that another point was provided, which would allow the poster to solve for "a" in "a(x - b)(x - c)(x - d)". But since that point is actually a fourth zero for a polynomial which can have only three, there must be something wrong with the exercise. Eliz.
morson said: You have told us that there are 4 zeros to a cubic... Click to expand... Good catch! I only noticed that another point was provided, which would allow the poster to solve for "a" in "a(x - b)(x - c)(x - d)". But since that point is actually a fourth zero for a polynomial which can have only three, there must be something wrong with the exercise. Eliz.