Find all the real and imaginary zeros for the following polynomial T(b)=245b3−161b2−17b+5

Answer 1

Step 1:

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.

Step 2:

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

p/q	245	-161	-17	5
1	245	84	67	72
-1	245	-406	389	-384
-1/5	245	-210	25	0

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

Since the depressed polynomial 245b² - 210b + 25 is a 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

Apply zero product property.

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

b = 5/7 and b = 1/7

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

There are no imaginary zeros.

Solution:

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

There are no imaginary zeros.