Factor or simplify a^4 + a^3 + a^2 + a + 1, and say whether it factors over the rationals and over the reals.

Step-by-step solution

1. 1

   Recognise the geometric series. The five terms are consecutive powers $a^0$ through $a^4$ with ratio $a$, so the finite geometric sum formula applies:

   $a^4+a^3+a^2+a+1=\frac{a^5-1}{a-1},\qquad a\neq1$

   Equivalently, multiplying the expression by $(a-1)$ telescopes to $a^5-1$. At $a=1$ the value is simply $5$.

2. 2

   Rule out rational factors. By the Rational Root Theorem any rational root must divide the constant term $1$, so only $a=\pm1$ are candidates: $p(1)=5$ and $p(-1)=1-1+1-1+1=1$, neither zero. This polynomial is the fifth cyclotomic polynomial $\Phi_5$, and it is indeed irreducible over the rationals — its roots are the four primitive fifth roots of unity.

3. 3

   Do not stop there: it does factor over the reals. A common error is to conclude "irreducible" full stop. Every real quartic factors into two real quadratics, and here the factorisation is explicit. Divide through by $a^2$ (valid for $a\neq0$) and group symmetrically:

   $a^2+a+1+\frac1a+\frac{1}{a^2}=\left(a^2+\frac{1}{a^2}\right)+\left(a+\frac1a\right)+1$

4. 4

   Substitute t = a + 1/a. Since $a^2+\frac{1}{a^2}=t^2-2$, the bracket becomes

   $t^2-2+t+1=t^2+t-1$

   whose roots are $t=\dfrac{-1\pm\sqrt5}{2}$ — the golden ratio conjugates. Each value of $t$ corresponds to the quadratic $a^2-ta+1=0$.

5. 5

   Write the real factorisation.

   $a^4+a^3+a^2+a+1=\left(a^2+\frac{1+\sqrt5}{2}a+1\right)\left(a^2+\frac{1-\sqrt5}{2}a+1\right)$

   Both factors are irreducible over the reals: their discriminants are $t^2-4$ with $t\approx0.618$ and $t\approx-1.618$, giving $-3.618$ and $-1.382$, both negative.

6. 6

   Note the consequence and verify. Because both quadratic factors are always positive, the quartic is positive for every real $a$ — its minimum over $[-8,8]$ is about $0.6736$, never zero. Numerically, the original expression and the two-quadratic product agree at $80$ random values of $a$ in $[-4,4]$ to machine precision ✓, and the geometric-series form matches at $a=1.3,\,2.7,\,-0.4,\,-3.1$ ✓.