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.

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