Factor 21x^3 - 49x^2 - 18x + 42 completely, and find its roots.

Step-by-step solution

1. 1

   Try grouping before hunting for roots. A four-term cubic is the classic candidate for factoring in pairs. Split it down the middle:

   $\left(21x^{3}-49x^{2}\right)+\left(-18x+42\right).$

   Testing integers such as $x=2$ (giving $-22$) and $x=3$ (giving $114$) finds nothing, which is why a root-first approach stalls here — the only rational root is the fraction $\tfrac73$, easy to overlook among the many candidates $\pm\frac{p}{q}$ with $p\mid42$, $q\mid21$.

2. 2

   Factor each pair. From the first pair take out $7x^{2}$, and from the second take out $-6$:

   $21x^{3}-49x^{2}=7x^{2}(3x-7),\qquad -18x+42=-6(3x-7).$

   The sign matters: pulling out $-6$ rather than $6$ is what makes the second bracket $3x-7$ instead of $7-3x$.

3. 3

   Extract the common binomial. Both pieces now share the factor $(3x-7)$:

   $7x^{2}(3x-7)-6(3x-7)=(3x-7)\left(7x^{2}-6\right).$

   Expanding back confirms it: $21x^{3}-49x^{2}-18x+42$ ✓.

4. 4

   Check whether the quadratic factors further. Over the rationals, $7x^{2}-6$ is irreducible because $\tfrac67$ is not the square of a rational. Over the reals it does split:

   $7x^{2}-6=7\left(x-\sqrt{\tfrac67}\right)\left(x+\sqrt{\tfrac67}\right).$

   So the complete rational factorisation is $(3x-7)(7x^{2}-6)$.

5. 5

   Read off the roots and verify. Setting each factor to zero:

   $x=\frac{7}{3}\approx2.3333,\qquad x=\pm\sqrt{\frac67}=\pm\frac{\sqrt{42}}{7}\approx\pm0.92582.$

   Substituting each into the cubic returns $0$ to within $10^{-14}$ ✓, and Vieta agrees: the roots sum to $\tfrac73+0=\tfrac73=\tfrac{49}{21}$ ✓.