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Mathematics 15 Online
OpenStudy (anonymous):

use the rational zeros theorem to write a list of all potential rational zeros f(x) =13x3+23x2+2x-26

OpenStudy (mathstudent55):

a0 = -26 = p a3 = 13 = q List all factor of p and all factors of q. The potential zeros are all x = p/q and x = -p/q

OpenStudy (anonymous):

what is a0 and a3 ?

OpenStudy (mathstudent55):

\(\large f(x) = a_nx^n + a_{n - 1}x^{n - 1} + ... + a_3x^3 + a_2x^2 + a_1x + a_0\) \(a_3\) is the leading coefficient, the coefficient of the term of highest degree. \(a_0\) is the constant term, the term with no x.

OpenStudy (mathstudent55):

Compare the general form above with what you have: \(f(x) = 13x^3+23x^2+2x-26\) That means that for you, \(a_3 = 13\), and \(a_0 = -26\).

OpenStudy (mathstudent55):

Now find all the factors of \(a_0 = -26\) and all the factors of \(a_3 = 13\).

OpenStudy (anonymous):

so is it -/+1, -/+ 1/3, -/+ 2,-/+ 13, -/+ 26 ??

OpenStudy (mathstudent55):

Keep the factors of the two numbers separate, because the factors of a0 go in the numerator, and the factors of a3 go in the denominator. The factors of a0 = -26 are 1, 2, 13 The factors or a3 = 13 are 1, 13 The possible zeros are all combinations of p/q, both positive and negative. Remember, p comes from the constant term,and q comes from the leading coefficient. These are the possibilities: 1/1, 1/13, 2/1, 2/13, 13/1, 13/13 We can reduce to: 1, 1/13, 2, 2/13, 13, 1 Now eliminate repeats: 1, 1/13, 2, 2/13, 13 Now add the negatives: 1, -1, 1/13, -1/13, 2, -2, 2/13, -2/13, 13, -13

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