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Take the equation $ax^{2} + bx + c = 3x^{2} + 4x + 53$.

Why is it always true that $a = 3, b = 4$ and $c = 53$?

I've seen many examples like this where the coefficients are equated, and was just wondering why that is always true.

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    $\begingroup$ What do you mean by $ax^2+bx+c=3x^2+4x+53$? $\endgroup$ – Ofir Schnabel Jun 30 '15 at 9:43
  • $\begingroup$ That was just arbitrary. I meant why does a = d, b = e and d = f when $ax^{2}+bx+c = dx^{2} + ex = f$ $\endgroup$ – coolcheetah Jun 30 '15 at 9:44
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    $\begingroup$ If you have two elephants and 3 rhino does that equal 0 elephants and 1 rhino? $\endgroup$ – Chinny84 Jun 30 '15 at 9:45
  • $\begingroup$ So we can take also $3=8$ if we are in $\mathbb{Z}_5$. $\endgroup$ – Ofir Schnabel Jun 30 '15 at 9:45
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    $\begingroup$ The question as posed is missing the very important words "for all $x$." Without them, you do not know that all coefficients are equal. You could very well have $a=4\neq 3$, $b=4$, $c=52\neq 53$, provided that $x=1$ or $x=-1$. Or to put it another way, if $a\neq 3$ then you have a quadratic equation to solve for $x$ rather than a formula that is true for all $x$; and if $a=3$ but $b\neq 4$ you have a linear equation with a unique solution. $\endgroup$ – David K Jun 30 '15 at 12:35
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Suppose that $ax^{2} + bx + c = 3x^{2} + 4x + 53$ for all $x$, or $ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ for all $x$.

$ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ is a polynomial with at most degree $2$ so by the fundamental theorem of algebra it has at most $2$ roots if it is not the zero polynomial.

But $ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ has infintely many roots (it is zero for every value of $x$), so it must be the zero polynomial. Then we have $a=3$, $b=4$ and $c=53$.

$$ax^{2} + bx + c - (3x^{2} + 4x + 53) = (a-3)x^2+(b-4)x+(c-53)$$

Every coefficient must be equal to zero for it to be the zero polynomial, thus $a-3=0$, and $a=3$. Similiarly, one has $b=4$ and $c=53$.

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  • $\begingroup$ Why does it have infinitely many roots? $\endgroup$ – coolcheetah Jun 30 '15 at 9:46
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    $\begingroup$ Because it is zero everywhere. $ax^{2} + bx + c - 3x^{2} + 4x + 53 = 0$ for all $x$. $\endgroup$ – wythagoras Jun 30 '15 at 9:48
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    $\begingroup$ Not that it changes things much, but haven't you made a sign mistake in Line 2? $\endgroup$ – coolcheetah Jun 30 '15 at 9:51
  • $\begingroup$ @SR1 I forgot to put brackets. $\endgroup$ – wythagoras Jun 30 '15 at 9:53
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    $\begingroup$ $ax^{2} + bx + c = 3x^{2} + 4x + 53$, thus $ax^{2} + bx + c - (3x^{2} + 4x + 53)= (3x^{2} + 4x + 53) - (3x^{2} + 4x + 53)$, thus $ax^{2} + bx + c - (3x^{2} + 4x + 53)=0$. This holds for all $x$. Thus there are infinitely many roots. Note that $a$ is a root of $P(x)$ if $P(a)=0$. $\endgroup$ – wythagoras Jun 30 '15 at 10:00
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The equation is valid for all $x$ take $x=0$ to get $c=53$. The equation now rewrites as follows $ax^2+bx=3x^2+4x$ and again this is true for all $x$ especially for say $x=-{4\over 3}$ and the equation now looks like ${16a\over 9}-{4b\over 3}=0$ and for $x=1$ we get $a+b=7$ the solution of the system in $(a,b)$ is $(3,4)$

In general if two polynomial are such as $a_nx^n+\cdots+a_0=b_nx^n+\cdots+b_0$ for all x this means $(a_n-b_n)x^n+\cdots+(a_0-b_0)$ has an infinity of solutions while it has degree $n$ and therefore all its coefficients are $0$ which translates into $a_i=b_i\,\forall i$

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  • $\begingroup$ Thanks. Exactly what I wanted. But why does the last equation have an infinite number of solutions? $\endgroup$ – coolcheetah Jun 30 '15 at 9:59
  • $\begingroup$ True. Thank you for your time. $\endgroup$ – coolcheetah Jun 30 '15 at 10:04
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Here's a cute way of proving it I don't think has been mentioned yet.

If you agree two polynomials are equal, you should also agree that their derivatives are equal.

Let $f(x) = ax^2 + bx + c = 3x^2+4x+53$.

Taking the derivative of $f$, we get:

$f'(x) = 2ax + b = 6x + 4$.

Taking the second derivative, we get:

$f''(x) = 2a = 6$.

Working backwards, we can solve for each variable in turn. The last equation gives $a = 3$. Going to the one before it, we can replace $2ax$ with $6x$ and subtract $6x$ from both sides to get $b=4$. And finally, we go back to the original to subtract $3x^2 + 4x$ from both sides to get $c=53$.

It's not the most efficient way, but it doesn't require any evaluation of the polynomial or talk of its roots.

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    $\begingroup$ You can use mathematical induction for a rigorous proof. $\endgroup$ – lovetl2002 Jul 15 '20 at 5:50
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The set $\{1,x,x^2,...\}$ is a linear independent set in the vector space of real valued polynomials ($+$ and $\cdot$ defined in the obvious way). Then, by definition of linear independence, $$ \sum_{i=0}^n a_ix^i = \sum_{j=0}^m b_jx^j \iff a_i = b_j \quad \forall i,j<n=m. $$

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To try and give a more general answer to why $a=d$, $b=e$ and $c=f$ given that $ax^{2} + bx + c = dx^{2} + ex + f$.

Rearranging gives $ax^{2} + bx + c - (dx^{2} + ex + f) = 0$, combining factors of $x$ gives $(a-d)x^{2} + (b-e)x + (c-f) = 0$.

It then follows that each of $(a-d)x^{2}$, $(b-e)x$ and $(c-f)$ must equal $0$.

If $x$ can have any value then $a - d = 0$ therefore $a = d$ with the same for $b = e$ and $c = f$.

It should be noted that values can be found that mean $a$ and $d$, $b$ and $e$, and $c$ and $f$ are not equal for specific values of $x$. Such as $x$ = $0$, which allows $a$, $b$, $d$ and $e$ to have any value.

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