a prime polynomial cannot be written as a product of lower - degree polynomials. which polynomial is…

Explanation:

Step1: Try factoring $8x^{2}-10x - 3$

We use the AC - method. For $ax^{2}+bx + c=8x^{2}-10x - 3$, $a = 8$, $b=-10$, $c=-3$, and $ac=8\times(-3)=-24$. We need two numbers that multiply to $-24$ and add up to $-10$. The numbers are $-12$ and $2$. Then $8x^{2}-10x - 3=8x^{2}-12x+2x - 3 = 4x(2x - 3)+1(2x - 3)=(4x + 1)(2x - 3)$.

Step2: Try factoring $8x^{2}+2x - 3$

For $8x^{2}+2x - 3$, $a = 8$, $b = 2$, $c=-3$, $ac=8\times(-3)=-24$. The numbers that multiply to $-24$ and add up to $2$ are $6$ and $-4$. Then $8x^{2}+2x - 3=8x^{2}+6x-4x - 3=2x(4x + 3)-(4x + 3)=(2x - 1)(4x + 3)$.

Step3: Try factoring $8x^{2}-6x - 3$

For $8x^{2}-6x - 3$, $a = 8$, $b=-6$, $c=-3$, $ac=8\times(-3)=-24$. We try to find two numbers that multiply to $-24$ and add up to $-6$. There are no two such integers. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $8x^{2}-6x - 3$, where $a = 8$, $b=-6$, $c=-3$, we have $b^{2}-4ac=(-6)^{2}-4\times8\times(-3)=36 + 96=132$. Since $\sqrt{132}$ is not a rational number, and we are looking for factoring over the integers, $8x^{2}-6x - 3$ cannot be factored into lower - degree polynomials with integer coefficients.

Step4: Try factoring $8x^{2}+23x - 3$

For $8x^{2}+23x - 3$, $a = 8$, $b = 23$, $c=-3$, $ac=8\times(-3)=-24$. The numbers that multiply to $-24$ and add up to $23$ are $24$ and $-1$. Then $8x^{2}+23x - 3=8x^{2}+24x-x - 3=8x(x + 3)-(x + 3)=(8x - 1)(x + 3)$.

Answer:

$8x^{2}-6x - 3$