MATHalino

Because the first term is $2x^{2}$, we are certain that the factorization takes the form
\[
(2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)
\]Taking note that the factors of $6$ are $1\times6$ and $2\times3$, we now do some deductions.

• Since the third term $-6$ is negative, the two factors will take opposite signs of constants.
• It is certainly not $(2x\square6)(x\square1)$. Otherwise, the first factor is still factorable by GCF as $2(x\square3)$. But we cannot take out $2$ via GCF from the original trinomial.
• It is also certainly not $(2x\square1)(x\square6)$, because the outer terms multiply to $(2x)(6)=12x$, which is too large for the middle term $-x$ of the original trinomial. Also, notice that the inner terms multiply to $1(x)=x$, which is too small to reduce $12x$ to $-x$.

This rules out $6=1\times6$. Hence, we are left with $2\times3$ as the factors of $6$.

• It is certainly not $(2x\square2)(x\square3)$ for the same reason as $(2x\square6)(x\square1)$.

Hence, we are left with $(2x\square3)(x\square2)$. The outer terms multiply to $2x(2)=4x$, while the inner terms multiply to $3(x)=3x$.
\[
(\overbrace{2x\square\underbrace{3)(x}_{3x}\square2}^{4x})
\] Because we need to get the middle term $-x$, and because the two factors take opposite signs, the only possibility is that $4x$ must be negative and $3x$ must be positive.
\[
-4x+3x=-x
\] leading to
\[
(\overbrace{2x\square\underbrace{3)(x}_{3x}\square2}^{-4x})
\] Therefore, the second square must be negative, and the first square must be positive.
\[
(\overbrace{2x+\underbrace{3)(x}_{3x}-2}^{-4x})
\] Therefore, $2x^{2}-x-6=\boxed{(2x+3)(x-2)}$. Take note that all of these deductions are happening mentally.

Because the first term is $4x^{2}$, one possible combination is
\[
(2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)
\] Because the leading coefficient of each factor is $2$, an even number, we are sure that the outer and inner terms are even, and consequently the resulting middle term is even. This is a contradiction, because the middle term of the original polynomial is $-19x$, which is an odd number. Hence, we reject this one and embrace the remaining possible combination
\[
(4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)
\] Because the middle term is negative and the last term is positive, each of the factors must take negative signs.
\[
(4x-\square)(x-\square)
\] The last term $21$ has the possible factors $1\times21$ and $7\times3$. We now do some deductions.

• Definitely, $(x-1)$ is not one of the factors because the sum of the coefficients of the terms of the trinomial is $4-19+21=6\neq0$. So, $(4x-21)(x-1)$ is out.
• Also, it is definitely not $(4x-1)(x-21)$, because the outer terms multiply to $4x(-21)=-84x$, which is too large.

Hence, we are left with $21=7\times3$.

• It is not $(4x-3)(x-7)$, because the outer terms multiply to $(4x)(-7)=-28x$, which is too large compared to the middle term $-19x$.

Therefore, we are left with the factorization $(4x-7)(x-3)$. A customary check of the middle term gives
\[
4x(-3)+(-7)(x)=-12x-7x=-19x
\] which checks. Therefore, $4x^{2}-19x+21=\boxed{(4x-7)(x-3)}$. Take note that all of these deductions are happening mentally.

This trinomial looks scary because of large coefficients. However, notice that the sum of the coefficients and constant terms is zero.
\[
45+79-124=0
\] Therefore, one of the factors is automatically $(x-1)$. The other one is $({\color{blue}{45}}x{\color{red}{\,+\,124}})$. Therefore,
\[
{\color{blue}{45}}x^{2}+79x{\color{red}{\,-\,124}}=\boxed{(x-1)({\color{blue}{45}}x{\color{red}{\,+\,124}})}
\] Observe carefully the colored numbers.

Observe that the sum of the coefficients of the first and last terms equals the middle term.
\[
10+(-27)=-17
\] Therefore, one of the factors is automatically $(x+1y)$. The other one is $({\color{blue}{10}}x{\color{red}{\,-\,27}}y)$. Therefore,
\[
{\color{blue}{10}}x^{2}-17xy{\color{red}{\,-\,27}}y^{2}=\boxed{(x+y)({\color{blue}{10}}x{\color{red}{\,-\,27}}y)}
\]Again, observe carefully the colored numbers.

The leading coefficient $12$ has factors $1\times12$, $2\times6$ and $3\times4$. The constant term $-12$ also has factors $1\times12$, $2\times6$ and $3\times4$, signs not included.

• We will eliminate $12=2\times6$ immediately, because both $2$ and $6$ are even but the middle term $7x$ is odd.
• We will eliminate $12=1\times12$ also, because the only possible combinations which will yield an odd middle term are $(x\square4)(12x\square3)$ and $(x\square12)(12x\square1)$.
• $(x\square4)(12x\square3)$ is out, because $(12x\square3)$ is still factorable by GCF, but the original trinomial is not.
• $(x\square12)(12x\square1)$ is out, because the inner term gives $12(12x)=144x$, which is too large compared to the middle term $7x$.

So, we are left with $12=3\times4$.
\[
(3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)
\]Recall that the constant term $-12$ also has factors $1\times12$, $2\times6$ and $3\times4$, signs not included.

• Because the constant term is negative, the two factors take opposite signs.
• The factor with $4x$ can only take $\pm1$ and $\pm3$ as the constant term. Otherwise, this factor is still factorable via GCF, while the original trinomial is not, a contradiction.
• If the factor with $4x$ takes $\pm1$ as the constant term, then the factor with $3x$ takes $\pm12$ as the constant term, since $12=1\times12$. This is a contradiction, because the factor $(3x\square12)$ will now be factorable by GCF. Therefore, we eliminate $\pm1$.

Therefore, we are left with $\pm3$ as the constant term of the factor with $4x$, and $\pm4$ as the constant term of the factor with $3x$, since $12=3\times4$.
\[
(\overbrace{4x\square\underbrace{3)(3x}_{9x}\square4}^{16x})
\]To get the middle term $7x$, we must have $16x+(-9x)=7x$.
\[
(\overbrace{4x\square\underbrace{3)(3x}_{-9x}\square4}^{16x})
\]Therefore, the signs in the squares must be
\[
(\overbrace{4x-\underbrace{3)(3x}_{-9x}+4}^{16x})
\]Hence, $12x^{2}+7x-12=\boxed{(4x-3)(3x+4)}$, and we are done.