The polynomial T (b )= 245 b³ - 161b² - 17b + 5 = 0 \ \

Identify Rational Zeros  

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

Rational Root Theorem : If a rational number in simplest form p/q  is a root of the polynomial equation a_nxⁿ + a_n – 1x^{n – 1} + ... + a₁x + a₀ = 0, where p  is a factor of a₀ and q  is a factor if a_n. \ \

If p/q  is a rational zero, then p  is a factor of 5 and q  is a factor of 245. \ \

The possible values of p  are  ± 1 and ± 5. \ \

The possible values for q  are ± 1, ± 5, ± 7, ± 35, ± 49 and ± 245. \ \

By the Rational Roots Theorem, the only possible rational roots are, p/q =± 1, ± 1/5, ± 1/7, ± 1/35, ± 1/49,  ± 5, ± 5/7, ± 5/7 and ± 5/49. \ \

Make a table for the synthetic division and test possible real zeros.

Since T (-1/5) = 0, b  =  -1/5 is a zero. The depressed polynomial is  245b² - 210b + 25 = 0. \ \

Since the depressed polynomial of this zero, 245b² - 210b + 25, is quadratic, use the factorization to find the roots of the related quadratic equation 49b² - 42b + 5 = 0. \ \

49b² - 42b + 5 = 0 \ \

49b² - 35b - 7b + 5 = 0 \ \

7b (7b - 5) - 1(7b - 5) = 0 \ \

(7b - 5)(7b - 1) = 0 \ \

7b - 5 = 0 and 7b - 1 = 0 \ \

b = 5/7 and b = 1/7 \ \

The cubic polynomial has three real zeros are b = -1/5, b = 5/7 and b = 1/7. \ \

There are no imaginary zeros. \ \

Solution: \ \

The cubic polynomial has three real zeros are b = -1/5, b = 5/7 and b = 1/7. \ \

There are no imaginary zeros. \ \

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