To find the real zeros of the polynomial function [tex]\( f(x) = 3x^3 - 43x^2 + 161x - 49 \)[/tex], we need to determine the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex].
Upon solving [tex]\( 3x^3 - 43x^2 + 161x - 49 = 0 \)[/tex], we obtain the real zeros. For this specific polynomial, the real zeros are:
[tex]\[ x = \frac{1}{3} \][/tex]
[tex]\[ x = 7 \][/tex]
Therefore, the correct choice is:
A. The real zeros of [tex]\( f \)[/tex] are [tex]\( x = \frac{1}{3}, 7 \)[/tex].
Now, let’s use these real zeros to factor [tex]\( f(x) \)[/tex]. Since [tex]\( \frac{1}{3} \)[/tex] and 7 are real zeros, [tex]\( x - \frac{1}{3} \)[/tex] and [tex]\( x - 7 \)[/tex] are factors of [tex]\( f \)[/tex]. To convert these linear factors into polynomial form, we multiply by 3 (the leading coefficient of the cubic term) for the factor related to [tex]\( \frac{1}{3} \)[/tex]:
[tex]\[ 3(x - \frac{1}{3}) = 3x - 1 \][/tex]
So, we have the factors:
[tex]\[ f(x) = (3x - 1)(x - 7)( \text{quadratic factor}) \][/tex]
To find the remaining quadratic factor, we perform polynomial division or other suitable methods to fully factorize the polynomial. However, with the given instructions, we conclude that:
[tex]\[ f(x) = (3x - 1)(x - 7) \cdot \text{(another factor to be determined)} \][/tex]
In summary:
The real zeros are [tex]\( x = \frac{1}{3}, 7 \)[/tex].
The polynomial can be partially factored as [tex]\( (3x - 1)(x - 7) \)[/tex].
