Question

which polynomial is prime?
$x^3 + 3x^2 - 2x - 6$
$x^3 - 2x^2 + 3x - 6$
$4x^4 + 4x^3 - 2x - 2$
$2x^4 + x^3 - x + 2$

Explanation:

Step1: Factor first polynomial by grouping

Group terms: $(x^3+3x^2)+(-2x-6)$
Factor out GCFs: $x^2(x+3)-2(x+3)=(x+3)(x^2-2)$

Step2: Factor second polynomial by grouping

Group terms: $(x^3-2x^2)+(3x-6)$
Factor out GCFs: $x^2(x-2)+3(x-2)=(x-2)(x^2+3)$

Step3: Factor third polynomial by grouping

Group terms: $(4x^4+4x^3)+(-2x-2)$
Factor out GCFs: $4x^3(x+1)-2(x+1)=(x+1)(4x^3-2)$

Step4: Test fourth polynomial for factoring

Attempt grouping: $(2x^4+x^3)+(-x+2)=x^3(2x+1)-1(x-2)$. No common binomial factor. Test rational roots via Rational Root Theorem: possible roots are $\pm1,\pm2,\pm\frac{1}{2}$. Substitute each:

• $f(1)=2+1-1+2=4

eq0$

• $f(-1)=2-1+1+2=4

eq0$

• $f(2)=32+8-2+2=40

eq0$

• $f(-2)=32-8+2+2=28

eq0$

• $f(\frac{1}{2})=\frac{2}{16}+\frac{1}{8}-\frac{1}{2}+2=\frac{1}{8}+\frac{1}{8}-\frac{4}{8}+\frac{16}{8}=\frac{14}{8}

eq0$

• $f(-\frac{1}{2})=\frac{2}{16}-\frac{1}{8}+\frac{1}{2}+2=\frac{1}{8}-\frac{1}{8}+\frac{4}{8}+\frac{16}{8}=\frac{20}{8}

eq0$
No rational roots, so it cannot be factored over integers.

Answer:

$2x^4 + x^3 - x + 2$