Respuesta :
As already described above, the prime polynomial is that which can no longer be factored to yield a much simpler terms. From the given choices, that which cannot be factored is letter C.
A prime polynomial cannot be written as a product of lower-degree polynomials is [tex]=8x^{2} -6x-3= 0[/tex].
A polynomial with integer coefficients that cannot be factored into polynomials of lower degree , also with integer coefficients, is called an irreducible or prime polynomial .
According to the question,
- Function f(x) = [tex]8x^{2} - 10x-3 = 0[/tex]
[tex]=8x^{2} -4x -6x-3= 0\\=4x ( 2x-1) - 3 (2x -1) = 0 \\=(2x-1) (4x-3)[/tex]
So, polynomial f(x) is factorize in product of lower degree polynomial.
- Function f(x) [tex]=8x^{2} +2x-3= 0[/tex]
The factor for this function find using formula,
[tex]x = \frac{-b \pm\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]x = \frac{-2 \pm\sqrt{2^{2}-4.8.(-3) } }{2(8)}\\ \\x = \frac{-2 \pm\sqrt{100} }{16} \\\\x = \frac{-2+10}{16} \\\\x = \frac{8}{16} \\\\x = \pm\frac{1}{2}[/tex]
So, polynomial f(x) is factorize in product of lower degree polynomial.
- Function f(x) = [tex]=8x^{2} -6x-3= 0[/tex]
[tex]x = \frac{-6 \pm\sqrt{(-6)^{2}-4.8.(-3) } }{2(8)}\\ x = \frac{-6 \pm\sqrt{132} }{16} \\[/tex]
So, polynomial f(x) is not factorize in product of lower degree polynomial.
- Function f(x) = [tex]8x^{2} +23x-3 = 0[/tex]
[tex]x = \frac{-23\pm\sqrt{23^{2}-4.8.(-3) } }{2(8)}\\ \\x = \frac{-2 \pm\sqrt{625} }{16} \\\\x = \frac{-23+25}{16} \\\\x = \frac{2}{16} \\\\x = \pm\frac{1}{8}[/tex]
So, polynomial f(x) is factorize in product of lower degree polynomial
Hence, A prime polynomial cannot be written as a product of lower-degree polynomials is [tex]=8x^{2} -6x-3= 0[/tex]
For more information about Quadratic equation click the link given below.
https://brainly.com/question/11589380
