To prove ax2+bx+c≡ay2+by+c(bmodm) if a, b and c are integers and x≡y(bmodm).

Step by step solution to "To prove ax2+bx+c≡ay2+by+c(bmodm) if a, b and c..."

High School
Algebra
Algebra II
Polynomial factorization

Cem Hayes

Answered question

2021-04-11

To prove $a x^{2} + b x + c \equiv a y^{2} + b y + c (b \mod m)$ if a, b and c are integers and $x \equiv y (b \mod m)$ .

Answer & Explanation

pattererX

Skilled2021-04-13Added 95 answers

Formula used:
Concept of congruence.
x is said to be in congruence with yif $x - y$ is divisible by m. Thus,
$x \equiv y (b \mod m)$
Calculation:
Consider that a, b, and care integers also $x \equiv y (b \mod m)$ .
Since $x \equiv y (b \mod m)$ , and $x - y$ is divisible by m. Therefore,
$a (x - y) (x + y) a n d b (x - y)$ is divisible by m. Further,
$a (x^{2} - y^{2})$ and $b (x - y)$ is divisible b m. Further,
$a (x^{2} - y^{2}) + b (x - y) + c - c$ is divisible by m.
$[a x^{2} + b x + c] - [a y^{2} + b y + c]$ is divisible by m.
$[a x^{2} + b x + c] \equiv [a^{2} + b y + c] (b \mod m)$
Therefore, it is evident that if $x \equiv y (b \mod m)$ then
$a x^{2} + b x + c \equiv a y^{2} + b y + c (b \mod m)$ .

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