Sum of integer coefficients sum of which equals $2020$ of a quadratic polynomial that has 2 integer roots differing by $1$

54 Views Asked by At

We have a quadratic polynomial $ax^2+bx+c$ for $a,b,c \in \mathbb{Z}, a+b+c=2020$ with roots $k, l \in \mathbb{Z}, |k-l| = 1$. The task of the entire exercise is to find how many such polynomials exist. I am stuck at this particular step:

Solution written in the book this problem comes from says

The polynomial can be rewritten as $a(x-k)(x-l)$. Considering that the roots differ by $1$, the sum of our coefficients can be simply expressed as $a(k-1)(l-1)$, which is equal to $2020$.

But my question is, why can the sum of coeffecients be expressed as $a(k-1)(l-1)$ and how can that be proven? The authors haven't really given much information about this step.

Knowing this "formula", the rest of the exercise just comes down to prime factorization of $2020$ is not difficult to finish.

1

There are 1 best solutions below

0
On BEST ANSWER

$$ax^2+bx+c=a(x-k)(x-l)$$ $$a\cdot 1^2+b\cdot 1+c=a(1-k)(1-l) \tag{put x=1}$$ $$a+b+c=a(k-1)(l-1)$$

Hope it helps