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So this question essentially arose from setting:

$ax^2 + bx + c = (Ax + B)(Cx + D)$, where $A, B, C, D$ are integers.

This means $ax^2 + bx + c = ACx^2 + (BC + AD)x+ BD$. Now intuitively we've been thought in grade school that this means $a = AC$ , $b = (BC + AD)$ , and $c = BD$. However, I never really thought why 2 quadratics with different coefficients can't have the same graph (or same output for each real number input).

So my way of proving it was proof by contradiction.

Assume there exists $a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}$ within real numbers where $a_{2} \neq a_{1}$ or $b_{2} \neq b_{1}$ or $c_{2} \neq c_{1}$ such that for all $x$ within real numbers $a_{1}x^2 + b_{1}x + c_{1} = a_{2}x^2 + b_{2}x + c_{2}$.

This means $(a_{1} - a_{2})x^2 + (b_{1} - b_{2})x + (c_{1} - c_{2}) = 0$ for all real number inputs, meaning this new quadratic has infinitely many roots, which is contradiction since a quadratic only has at most 2 real roots.

Hence for all $a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}$ within real numbers where $a_{2} \neq a_{1}$ or $b_{2} \neq b_{1}$ or $c_{2} \neq c_{1}$ there exists $x$ within real numbers such that $a_{1}x^2 + b_{1}x + c_{1} \neq a_{2}x^2 + b_{2}x + c_{2}$.

Is my proof correct? I'm specifically not to confident if my final conclusion ("Hence for all ...") is enough to say that the coefficients must match. Since my conclusion basically says if any 1 of the coefficients of the 2 quadratics mismatch, then they won't have an equal output for some input x. Hence the only way for these 2 quadratics to have the same graph is for their coefficients to be equal (as in be the same equation). Not sure if that is implied or if I have to explicitly state it.

Also, I guess this argument extends for a polynomial of any degree, no? Kindly please let me know.

Bob Marley
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    As for why "$\left(a_1-a_2\right)x^2+\left(b_1-b_2\right)x+\left(c_1-c_2\right) = 0$ for all real number inputs" means that $a_1-a_2 = 0$ and $b_1-b_2 = 0$ and $c_1-c_2 = 0$, this question is related: Polynomial $p(x) = 0$ for all $x$ implies coefficients of polynomial are zero. – peterwhy Aug 02 '24 at 02:24
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    For a tedious mechanical proof, evaluate both quadratics at $x=-1,0,1$ and show that this implies all coefficients match. – Semiclassical Aug 02 '24 at 02:35
  • If the "it" in "would it not be a quadratic then?" means the function $\left(a_1-a_2\right)x^2+\left(b_1-b_2\right)x+\left(c_1-c_2\right)$, then right, it is the zero polynomial and not quadratic (and not linear). – peterwhy Aug 02 '24 at 02:57
  • Please review https://math.stackexchange.com/tags/solution-verification/info – D.W. Aug 02 '24 at 06:11
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    I'll say it in a slightly different way to avoid too much tedium. Let $a=a_1-a_2,b=b_1-b_2,c=c_1-c_2$. Then the condition for the two quadratics to agree for all $x$ is $ax^2+bx+c=0$. That would in particular have to be true when $x=0$, in which case the equation becomes $c=0$. Similarly, for $x=1,-1$ we get $a+b+c=0$ and $a-b+c=0$. But this means $a+b=a-b=0$, which requires $a=b=0$. Hence $a=b=c=0$, which amounts to all pairs of coefficients matching. – Semiclassical Aug 02 '24 at 07:28
  • @Semiclassical Oh I see. So I used the fact of number of roots of quadratic in my proof by contradiction, but you're proof used the assumption of working for all real number inputs and focused on x = -1, 0, 1 and solved for the coefficients of the difference equation/quadratic to all equal 0, which obviously means the coefficient of the 2 quadratics are also equal! Nice! – Bob Marley Aug 02 '24 at 22:24
  • @D.W. Just added the context. Kindly please help clarify my doubts. – Bob Marley Aug 02 '24 at 22:29
  • @peterwhy Hey I just added the specific thing I wanted to verify in my proof in my question statement (See paragraph "Is my proof correct? ...") Kindly please do help clarify my doubts. – Bob Marley Aug 02 '24 at 22:31
  • @Semiclassical Hey I just added the specific thing I wanted to verify in my proof in my question statement (See paragraph "Is my proof correct? ...") Kindly please do help clarify my doubts. – Bob Marley Aug 02 '24 at 22:31

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