Question:

Step 1: Analyze the polynomial $p(x) = x^4 + 3x^3 + 2(1-a)x^2 - 3ax + a^2 = 0$ for only real roots. Step 2: Attempt to factor the polynomial. Try $p(x) = (x^2 + bx + c)(x^2 + dx + e)$. Expanding and matching coefficients with $b+d=3$, $c+e+bd = 2(1-a)$, $be+cd = -3a$, $ce = a^2$. Step 3: Try $c = a$ and $e = a$ (since $ce = a^2$). Then $be + cd = a(b+d) = 3a$, but we need $-3a$, so try $c = a, e = a$ with $b+d = 3$ gives $a(b+d) = 3a \neq -3a$ unless $a = 0$. Step 4: Try $c = -a, e = -a$: $ce = a^2$ ✓. Then $be + cd = -a(b+d) = -3a$ ✓ (since $b+d=3$). And $c+e+bd = -2a + bd = 2(1-a)$, so $bd = 2$. With $b+d=3$ and $bd=2$: $b=1, d=2$ or $b=2, d=1$. Step 5: So $p(x) = (x^2 + x - a)(x^2 + 2x - a)$. For all roots to be real, both quadratics must have non-negative discriminants. Step 6: Discriminant of $x^2 + x - a \geq 0$: $1 + 4a \geq 0 \Rightarrow a \geq -\dfrac{1}{4}$. Step 7: Discriminant of $x^2 + 2x - a \geq 0$: $4 + 4a \geq 0 \Rightarrow a \geq -1$. Step 8: Combined condition: $a \geq -\dfrac{1}{4}$. Among the options, $a = 0$ and $a = 1$ satisfy this. $a = -1$ does not satisfy $a \geq -1/4$, and $a = 2$ gives discriminants $1+8=9>0$ and $4+8=12>0$, so $a=2$ also works. Step 9: Re-check $a=2$: $(x^2+x-2)(x^2+2x-2)=0$. $x^2+x-2=(x+2)(x-1)$, real roots ✓. $x^2+2x-2=0$, discriminant $= 4+8=12>0$, real roots ✓. So $a=2$ works. Step 10: Check $a=0$: $(x^2+x)(x^2+2x) = x(x+1)\cdot x(x+2)$, all real roots ✓. Check $a=1$: $(x^2+x-1)(x^2+2x-1)$, discriminants $1+4=5>0$ and $4+4=8>0$ ✓. The correct answers are $a=0$ and $a=1$ (options 2 and 3).