Question

which polynomial is prime?
\\( x^3 + 3x^2 + 2x + 6 \\)
\\( x^3 + 3x^2 - 2x - 6 \\)
\\( 10x^2 - 4x + 3x + 6 \\)
\\( 10x^2 - 10x + 6x - 6 \\)

Explanation:

Step1: Recall Prime Polynomial

A prime polynomial (or irreducible polynomial) cannot be factored into non - constant polynomials with integer coefficients. We will check each polynomial for factorability.

Step2: Analyze \(x^{3}+3x^{2}+2x + 6\)

Group the terms: \((x^{3}+3x^{2})+(2x + 6)=x^{2}(x + 3)+2(x + 3)=(x^{2}+2)(x + 3)\). So it is not prime.

Step3: Analyze \(x^{3}+3x^{2}-2x - 6\)

Group the terms: \((x^{3}+3x^{2})+(-2x - 6)=x^{2}(x + 3)-2(x + 3)=(x^{2}-2)(x + 3)\). So it is not prime.

Step4: Analyze \(10x^{2}-4x + 3x + 6\)

First, combine like terms: \(10x^{2}-x + 6\). Let's check if it can be factored. The discriminant of the quadratic \(ax^{2}+bx + c\) (here \(a = 10\), \(b=- 1\), \(c = 6\)) is \(D=b^{2}-4ac=(-1)^{2}-4\times10\times6=1 - 240=-239<0\). But we can also try factoring by grouping (original non - combined form: \(10x^{2}-4x+3x + 6=(10x^{2}-4x)+(3x + 6)=2x(5x - 2)+3(x + 2)\). There is no common binomial factor, but wait, we made a mistake in combining. Wait, the original polynomial before combining is \(10x^{2}-4x + 3x+6=10x^{2}-x + 6\). But let's go back to the original non - combined terms: \(10x^{2}-4x+3x + 6\). Wait, actually, \(10x^{2}-4x+3x + 6=10x^{2}-x + 6\) is incorrect. Wait, \(10x^{2}-4x+3x+6=10x^{2}+(-4x + 3x)+6=10x^{2}-x + 6\). But if we consider the original grouping: \(10x^{2}-4x+3x + 6=(10x^{2}+3x)+(-4x - 6)=x(10x + 3)-2(2x + 3)\). No common factor. Wait, no, let's re - check. Wait, the third polynomial is \(10x^{2}-4x + 3x+6=10x^{2}-x + 6\). But let's check the fourth polynomial first.

Step5: Analyze \(10x^{2}-10x+6x - 6\)

Combine like terms: \(10x^{2}-4x - 6\). Factor out a 2: \(2(5x^{2}-2x - 3)\). And \(5x^{2}-2x - 3=(5x + 3)(x - 1)\), so \(10x^{2}-10x + 6x-6=2(5x + 3)(x - 1)\). Not prime.

Wait, let's re - check the third polynomial: \(10x^{2}-4x + 3x+6=10x^{2}-x + 6\). Wait, maybe I made a mistake in the third polynomial's original form. Wait, the third polynomial is \(10x^{2}-4x + 3x+6\). Let's group as \((10x^{2}+3x)+(-4x - 6)=x(10x + 3)-2(2x + 3)\). There is no common factor. But wait, the second - degree polynomial \(10x^{2}-x + 6\): the product of \(a = 10\) and \(c = 6\) is \(60\). We need two numbers that multiply to \(60\) and add to \(-1\). The factors of \(60\) are \(1\) and \(60\), \(2\) and \(30\), \(3\) and \(20\), \(4\) and \(15\), \(5\) and \(12\), \(6\) and \(10\). None of these pairs add up to \(-1\). But wait, the first two polynomials are cubic and we factored them. The fourth is quadratic (after combining) and factored. Wait, the third polynomial: Wait, maybe I misread the third polynomial. Wait, the third polynomial is \(10x^{2}-4x + 3x+6\). Let's re - group: \(10x^{2}+3x-4x - 6=x(10x + 3)-2(2x + 3)\). No common factor. But wait, the second polynomial: Wait, no, let's check the cubic polynomials again. Wait, the first cubic: \(x^{3}+3x^{2}+2x + 6=(x^{3}+3x^{2})+(2x + 6)=x^{2}(x + 3)+2(x + 3)=(x^{2}+2)(x + 3)\). The second cubic: \(x^{3}+3x^{2}-2x - 6=(x^{3}+3x^{2})+(-2x - 6)=x^{2}(x + 3)-2(x + 3)=(x^{2}-2)(x + 3)\). The fourth quadratic (after combining \(10x^{2}-10x + 6x-6=10x^{2}-4x - 6 = 2(5x^{2}-2x - 3)=2(5x + 3)(x - 1)\)). The third polynomial: \(10x^{2}-4x+3x + 6=10x^{2}-x + 6\). Wait, maybe the third polynomial is \(10x^{2}-4x+3x + 6\) (original) and when we try to factor, we can't find two binomials \((ax + b)(cx + d)\) such that \(ac = 10\), \(bd=6\) and \(ad+bc=-4 + 3=-1\). Let's check all possibilities for \(10x^{2}-x + 6\):

If \(a = 10\), \(c = 1\), then \(bd = 6\). Possible pairs for \((b,d)\) are \((1,6),(2,3),(-1,-6),(-2,-3)\). Then \(ad+bc=10d + b\). For \((b = 1,d = 6)\): \(10\times6+1=61 eq - 1\). For \((b = 2,d = 3)\): \(10\times3+2=32 eq - 1\). For \((b=-1,d=-6)\): \(10\times(-6)+(-1)=-61 eq - 1\). For \((b=-2,d=-3)\): \(10\times(-3)+(-2)=-32 eq - 1\).

If \(a = 5\), \(c = 2\), then \(bd = 6\). Possible pairs for \((b,d)\): \((1,6),(2,3),(-1,-6),(-2,-3)\). Then \(ad+bc=5d+2b\). For \((b = 1,d = 6)\): \(5\times6+2\times1=32 eq - 1\). For \((b = 2,d = 3)\): \(5\times3+2\times2=19 eq - 1\). For \((b=-1,d=-6)\): \(5\times(-6)+2\times(-1)=-32 eq - 1\). For \((b=-2,d=-3)\): \(5\times(-3)+2\times(-2)=-19 eq - 1\).

So \(10x^{2}-4x + 3x + 6=10x^{2}-x + 6\) is prime? Wait, no, wait the original third polynomial is \(10x^{2}-4x + 3x+6\). Wait, maybe I made a mistake in the first two cubics. Wait, the first cubic: \(x^{3}+3x^{2}+2x + 6\) factors as \((x + 3)(x^{2}+2)\), the second cubic: \(x^{3}+3x^{2}-2x - 6=(x + 3)(x^{2}-2)\), the fourth quadratic: \(10x^{2}-10x + 6x-6=10x^{2}-4x - 6=2(5x^{2}-2x - 3)=2(5x + 3)(x - 1)\), and the third quadratic: \(10x^{2}-4x+3x + 6=10x^{2}-x + 6\) which cannot be factored with integer coefficients. Wait, but let's check the problem again. Wait, the third option is \(10x^{2}-4x + 3x + 6\) (maybe a typo, but assuming the given form). Wait, no, maybe the third polynomial is \(10x^{2}-4x + 3x + 6\) and the fourth is \(10x^{2}-10x + 6x-6\). After checking all, the prime polynomial is \(10x^{2}-4x + 3x + 6\) (the third option). Wait, no, wait the third polynomial: let's re - check the combination. \(10x^{2}-4x+3x + 6=10x^{2}-x + 6\). And since we can't factor it into two non - constant polynomials with integer coefficients, it is prime. The other polynomials can be factored:

- \(x^{3}+3x^{2}+2x + 6=(x + 3)(x^{2}+2)\)
- \(x^{3}+3x^{2}-2x - 6=(x + 3)(x^{2}-2)\)
- \(10x^{2}-10x + 6x-6=2(5x + 3)(x - 1)\)

Answer:

The third option: \(10x^{2}-4x + 3x + 6\) (the one with the third circle)