Question

which polynomial is prime?
7x² - 35x + 2x - 10
9x³ + 11x² + 3x - 33
10x³ - 15x² + 8x - 12
12x⁴ + 42x² + 4x² + 14

Explanation:

Step1: Factor first polynomial

Group terms: \((7x^2 - 35x)+(2x - 10)\). Factor out GCF: \(7x(x - 5)+2(x - 5)=(7x + 2)(x - 5)\). Not prime.

Step2: Factor second polynomial

Group terms: \((9x^3 + 11x^2)+(3x - 33)\). Factor out GCF: \(x^2(9x + 11)+3(x - 11)\). No common binomial factor, but wait, re - check grouping. Wait, maybe wrong grouping. Let's try \((9x^3+3x)+(11x^2 - 33)\). Factor: \(3x(3x^2 + 1)+11(x^2 - 3)\). Still no common factor? Wait, no, original second polynomial: \(9x^3+11x^2 + 3x - 33\). Wait, maybe I made a mistake. Wait, let's check third polynomial.

Step3: Factor third polynomial

Group terms: \((10x^3 - 15x^2)+(8x - 12)\). Factor GCF: \(5x^2(2x - 3)+4(2x - 3)=(5x^2 + 4)(2x - 3)\). Not prime.

Step4: Factor fourth polynomial (assuming a typo, maybe \(12x^4+42x^2 + 4x^2+14\) is \(12x^4 + 46x^2+14\)? No, wait, original is \(12x^4+42x^2 + 4x^2+14\), combine like terms: \(12x^4+46x^2 + 14\). Factor out 2: \(2(6x^4 + 23x^2+7)\). Let's factor \(6x^4 + 23x^2+7\). Let \(y = x^2\), then \(6y^2+23y + 7\). Discriminant: \(23^2-4\times6\times7=529 - 168 = 361=19^2\). Roots: \(\frac{-23\pm19}{12}\). \(\frac{-23 + 19}{12}=\frac{-4}{12}=-\frac{1}{3}\), \(\frac{-23 - 19}{12}=\frac{-42}{12}=-\frac{7}{2}\). So \(6y^2+23y + 7=6(y+\frac{1}{3})(y+\frac{7}{2})=(3y + 1)(2y + 7)=(3x^2 + 1)(2x^2+7)\). So \(12x^4+46x^2 + 14=2(3x^2 + 1)(2x^2+7)\). Wait, but the second polynomial: \(9x^3+11x^2 + 3x - 33\). Let's try grouping again: \((9x^3+3x)+(11x^2 - 33)=3x(3x^2 + 1)+11(x^2 - 3)\). There is no common binomial factor, and it can't be factored into polynomials with integer coefficients. So the prime polynomial is \(9x^3 + 11x^2+3x - 33\).

Answer:

\(9x^3 + 11x^2+3x - 33\) (the second option: \(9x^3 + 11x^2+3x - 33\))