Factor (x - 1)(x - 2)(x + 3)(x + 6) - 5x^2 completely.

Step-by-step solution

1. 1

   Pair the brackets so the constants match. There are three ways to split four factors into two pairs; the useful one makes the constant terms agree. Since $(-1)(6)=-6$ and $(-2)(3)=-6$, pair the first with the fourth and the second with the third:

   $(x-1)(x+6)=x^{2}+5x-6,\qquad (x-2)(x+3)=x^{2}+x-6.$

   Choosing any other pairing gives two quadratics with different constants and no shared structure to exploit.

2. 2

   Introduce a substitution for the common part. Both quadratics contain $x^{2}+x-6$, because

   $x^{2}+5x-6=\left(x^{2}+x-6\right)+4x.$

   Setting $y=x^{2}+x-6$, the expression becomes

   $(y+4x)\,y-5x^{2}=y^{2}+4xy-5x^{2}.$

   This is now a homogeneous quadratic in the two quantities $y$ and $x$ — far easier than a general quartic.

3. 3

   Factor the quadratic in $y$. Look for two terms whose product is $-5x^{2}$ and whose sum is $4x$: these are $5x$ and $-x$. Hence

   $y^{2}+4xy-5x^{2}=(y+5x)(y-x).$

   The check is immediate: $(y+5x)(y-x)=y^{2}-xy+5xy-5x^{2}=y^{2}+4xy-5x^{2}$ ✓.

4. 4

   Substitute back. Replacing $y=x^{2}+x-6$ in each factor:

   $y+5x=x^{2}+x-6+5x=x^{2}+6x-6,\qquad y-x=x^{2}+x-6-x=x^{2}-6.$

   Therefore

   $(x-1)(x-2)(x+3)(x+6)-5x^{2}=\left(x^{2}+6x-6\right)\left(x^{2}-6\right).$

5. 5

   Check and note the roots. Numerical check at $x=1.7$: the original gives $-22.0499$ and the factored form gives $-22.0499$ ✓. Neither factor is reducible over the rationals: $x^{2}-6$ has roots $\pm\sqrt6$ and $x^{2}+6x-6$ has discriminant $36+24=60$, giving roots $-3\pm\sqrt{15}$. So all four roots are irrational, which is why no amount of rational-root testing would have found this factorisation.