The quadratic equation is $$a^2-9a+14=0$$ I know that the product of the last terms of the binomial for an equation equals the third term of the polynomial. Also, the sum of the products of those two numbers should be the middle (second) term of the polynomial.
But what two numbers multiply to give 14 and added together produce −9?
So here is a trick. The last term, $14$, put into your calculator as $\frac{14}{x}$. No see what what two numbers will give you $14$. You should see that the two numbers that you should pick have the property that if you mutiple them it will give you $14$ and if you add them they should give you your middle number $-9$. So, $-2 \times -7 =14$ and $-2 + -7 = -9$, meaning: $$a^2 -9a +14 = 0 \\ (a-2)(a-7) = 0 \rightarrow a = 2 \text{ and } a=7$$
To answer your question, there is one key property to understand when it comes to quadratics and numbers in general.
The equation is as follows:
$a^2 - 9a + 14 = 0$
Let us first look into the factors of the constant, $14$.
The factors of $14$ are as follows:
$$1, 2, 7, 14$$
It is important to note that in the case of quadratics, you can make use of the negative factors as well. For example, $7$ multiplied by $2$ gives $14$, but so does $-7$ multiplied by $-2$. A positive number multiplied by a positive number always gives a positive number as a result. In addition, a negative number multiplied by a negative number also gives a positive number as a result.
In this case, we can see that the second term, $-9a$, fits nicely with the two factors $-7$ and $-2$. $-7$ and $-2$ multiply to give $14$, and when added together, give $-9$ (since $-7 + (-2) = -7 - 2 = -9$).
Therefore, we can factor the equation as follows:
$$a^2 - 9a + 14 = 0$$ $$a^2 - 7a - 2a + 14 = 0$$ $$a(a - 7) - 2(a - 7) = 0$$ $$(a - 2)(a - 7) = 0$$
From here, it is relatively easy to find the values of $a$ which satisfy the original equation (simply equate each factor to $0$ and solve the resulting equation).
I hope this helps! I'm still new to MSE, so I would really appreciate any and all feedback. Thank you!
Here is a trick that is about 2000 years old. It's actually kind of advanced, but also kind of interesting (I think).
$$x^2-9x+14=(x-r)(x-s)=x^2 -(r+s)x + rs$$
So you are looking for numbers $r$ and $s$ such that
$$\text{$r+s = 9$ and $rs = 14$}$$
Let $r = \dfrac 92 - t$ and $s = \dfrac 92 + t$.
I know this doesn't make any sense, but, stay with me here.
Clearly $r + s = 9$. So we have already solved that part.
Next we work on $rs = 14$.
\begin{align} rs &= 14 \\ \left(\dfrac 92 - t \right)\left(\dfrac 92 + t \right) &= 14 \\ \frac{81}{4} - t^2 &= 14 \\ 81 - 4t^2 &= 56 \\ 4t^2 &= 25 \\ 2t &= \pm 5 \\ t &= \pm \frac 52 \\ \end{align}
So $\{r,s\} = \left\{ \dfrac 92 + \dfrac 52, \dfrac 92 - \dfrac 52 \right\} = \left\{ 7, 2 \right\}$
