Factor 3x^5 + 9x^4 + 4x^3 - 9x^2 - 15x - 10 completely, and list its real roots.

Step-by-step solution

1. 1

   Search for a rational root. Possible rational roots are $\pm\frac{p}{q}$ with $p\mid 10$ and $q\mid 3$: $\pm1,\pm2,\pm5,\pm10,\pm\tfrac13,\pm\tfrac23,\pm\tfrac53,\pm\tfrac{10}{3}$. Testing $x=-2$:

   $3(-32)+9(16)+4(-8)-9(4)-15(-2)-10=-96+144-32-36+30-10=0,$

   so $x=-2$ is a root and $(x+2)$ is a factor. Ad-hoc grouping fails on this polynomial, which is why the root search is the right opening move.

2. 2

   Divide out $(x+2)$ by synthetic division. Bringing down the coefficients $3,\;9,\;4,\;-9,\;-15,\;-10$ and multiplying by $-2$ at each step gives the row $3,\;3,\;-2,\;-5,\;-5$ with remainder $0$:

   $3x^{5}+9x^{4}+4x^{3}-9x^{2}-15x-10=(x+2)\left(3x^{4}+3x^{3}-2x^{2}-5x-5\right).$

   A zero remainder is the confirmation that the root was correct.

3. 3

   Show the quartic has no rational root, then factor it into quadratics anyway. Testing $\pm1,\pm5,\pm\tfrac13,\pm\tfrac53$ in $3x^{4}+3x^{3}-2x^{2}-5x-5$ gives nothing zero (for instance the value at $x=-1$ is $-2$), so no further linear factor exists over $\mathbb{Q}$. Instead set

   $3x^{4}+3x^{3}-2x^{2}-5x-5=(3x^{2}+ax+b)(x^{2}+cx+d).$

4. 4

   Match coefficients and solve the small system. Expanding the product gives

   $3x^{4}+(3c+a)x^{3}+(3d+ac+b)x^{2}+(ad+bc)x+bd,$

   so $3c+a=3$, $3d+ac+b=-2$, $ad+bc=-5$ and $bd=-5$. Trying $b=-5$ forces $d=1$, and then $c=1$, $a=0$ satisfies all four equations. Hence

   $3x^{4}+3x^{3}-2x^{2}-5x-5=(3x^{2}-5)(x^{2}+x+1).$

   Check by expanding: $3x^{4}+3x^{3}+3x^{2}-5x^{2}-5x-5=3x^{4}+3x^{3}-2x^{2}-5x-5$ ✓.

5. 5

   Assemble the full factorisation and read off the real roots.

   $3x^{5}+9x^{4}+4x^{3}-9x^{2}-15x-10=(x+2)(3x^{2}-5)(x^{2}+x+1).$

   Setting each factor to zero: $x=-2$; $3x^{2}=5$ gives $x=\pm\sqrt{\tfrac53}=\pm\tfrac{\sqrt{15}}{3}\approx\pm1.290994$; and $x^{2}+x+1$ has discriminant $1-4=-3<0$, contributing no real root. So there are exactly three real roots.

6. 6

   Confirm numerically. Solving the quintic numerically returns $-2$, $\pm1.2909944$, and the complex pair $-0.5\pm0.8660254i$ — precisely $-\tfrac12\pm\tfrac{\sqrt3}{2}i$, the primitive cube roots of unity, as $x^{2}+x+1$ predicts ✓.