which polynomial is prime?\n$x^3 + 3x^2 - 2x - 6$\n$x^3 - 2x^2 + 3x - 6$\n$4x^4 + 4x^3 - 2x - 2$\n$2x^4 +…
which polynomial is prime?\n$x^3 + 3x^2 - 2x - 6$\n$x^3 - 2x^2 + 3x - 6$\n$4x^4 + 4x^3 - 2x - 2$\n$2x^4 + x^3 - x + 2$
Answer
Explanation:
Step1: Recall Prime Polynomial Definition
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}-2x^{2}+3x - 6)
Group the terms: ((x^{3}-2x^{2})+(3x - 6)=x^{2}(x - 2)+3(x - 2)=(x^{2}+3)(x - 2)). So it is not prime.
Step4: Analyze (4x^{4}+4x^{3}-2x - 2)
Group the terms: ((4x^{4}+4x^{3})+(-2x - 2)=4x^{3}(x + 1)-2(x + 1)=(4x^{3}-2)(x + 1)=2(2x^{3}-1)(x + 1)). So it is not prime.
Step5: Analyze (2x^{4}+x^{3}-x + 2)
Try to find factors. Let's use the rational root theorem. The possible rational roots are (\pm1,\pm2,\pm\frac{1}{2}).
- For (x = 1): (2(1)^{4}+(1)^{3}-1 + 2=2 + 1-1 + 2=4\neq0)
- For (x=-1): (2(-1)^{4}+(-1)^{3}-(-1)+2=2 - 1 + 1+2=4\neq0)
- For (x = 2): (2(16)+8-2 + 2=32 + 8-2 + 2=40\neq0)
- For (x=-2): (2(16)+(-8)-(-2)+2=32-8 + 2+2=28\neq0)
- For (x=\frac{1}{2}): (2(\frac{1}{16})+\frac{1}{8}-\frac{1}{2}+2=\frac{1}{8}+\frac{1}{8}-\frac{4}{8}+\frac{16}{8}=\frac{1 + 1-4 + 16}{8}=\frac{14}{8}=\frac{7}{4}\neq0)
- For (x=-\frac{1}{2}): (2(\frac{1}{16})+(-\frac{1}{8})-(-\frac{1}{2})+2=\frac{1}{8}-\frac{1}{8}+\frac{1}{2}+2=\frac{5}{2}\neq0)
Since we can't find a linear factor and we also can't factor it by grouping (as we tried earlier methods), and we can't express it as a product of two non - constant polynomials with integer coefficients, (2x^{4}+x^{3}-x + 2) is a prime polynomial.