Use polynomial identities, factors, and unknown coefficients

The wording of the question usually tells you the shortest way in:

• If it says two forms are equivalent for all values of $x$, match coefficients, starting with the one that gives you the unknown most directly, like the constant term in the example above.
• If it tells you $x+3$ is a factor, plug in $x=-3$, the input that makes that factor $0$.
• If it asks for a coefficient sum like $a+b+c$, find the polynomial's value at $1$ by choosing the $x$ that makes what's inside it equal $1$.
• If it asks which of several long answer choices is equivalent to one expression, with only $x$ and known numbers in them, graph overlap in Desmos may be faster than expanding.

Here's the first idea in general form. If

for all $x$, then

That only works because of the words for all values. An ordinary equation like $2x+1=7$ is true only at its solution, $x=3$, so there you solve for $x$ instead of matching coefficients.

The coefficient sum takes one more idea. If $P(t)=at^2+bt+c$, then plugging in $t=1$ turns every power of $t$ into $1$:

So a coefficient sum is $P(1)$. Now suppose the problem tells you that

for all $x$, and asks for $a+b+c$. You need $P(1)$, so you need the inside, $2x-3$, to equal $1$. That happens at $x=2$, because $2(2)-3=1$. Plug $x=2$ into the right side:

You never had to find $a$, $b$, or $c$ on their own. You're free to pick any $x$ here because the problem has already told you the identity holds for every $x$.

Line terms up by their power, not by where they sit in the expression. When a power doesn't appear, its coefficient is $0$, even though you can't see it:

That invisible $0x^2$ still counts. If the other side has an $x^2$ term, its coefficient has to equal $0$.

Sometimes one coefficient leaves you with two possible values. Say matching a coefficient gives $(t+2)^2=16$. Then $t+2=4$ or $t+2=-4$, so $t=2$ or $t=-6$. Don't pick one yet. Keep both, then check another coefficient to see which one makes the whole identity true.

Say $x+3$ is a factor of

When one factor of a product is $0$, the whole product is $0$. The factor $x+3$ is $0$ at $x=-3$, so $P(-3)$ has to be $0$ too. Plug in $-3$:

Set that equal to $0$:

so $m=-3$. In general, a factor $x-c$ tells you to plug in $x=c$, and a factor $x+c$ tells you to plug in $x=-c$.

When the question gives you a factor like this and asks for one coefficient, plugging in the zero is the quickest way. You could also build the full product. The other factor has to be $x-6$, because $3$ times its constant must give $-18$:

Multiplying it out gives the same $m=-3$. Building the product is worth it when the question needs the factors themselves, or when you can see them right away.

When an identity has unknown constants, matching coefficients is usually the shortest exact method. Graphing can mislead you here. One point where two graphs meet doesn't prove they're the same curve. And if you give an unknown constant one convenient value so you can graph it, the curves can match by coincidence.

Desmos does win on a different kind of question: one that asks which of several long answer choices is equivalent to an expression, where each choice uses only $x$ and fixed numbers. Type the original expression once, then add each choice and check for full-curve overlap, meaning the choice's graph sits exactly on top of the original everywhere. That compares finished expressions. It can't find an unknown coefficient for you.