After you distribute $x$ and $-3$, make sure to combine the $x^2$ terms together and the $x$ terms together so the expression looks like $x^3 + Ax^2 + Bx - 18$ for some $A$ and $B$ in terms of $t$.

Once your expanded expression is written as $x^3 + (t - 3)x^2 + (6 - 3t)x - 18$, compare it to $x^3 + 2x^2 + kx - 18$. What equations do you get by matching the coefficients of $x^2$ and $x$?

After solving for $t$ and then for $k$, remember the question asks for $t + k$, not just $t$ or $k$ alone.

After you have found numerical values for $t$ and $k$ by hand, type the first expression as y = (x^2 + t*x + 6)*(x - 3) with your value substituted for t, and the second as y = x^3 + 2*x^2 + k*x - 18 with your value substituted for k.

Look at the two graphs: if they lie exactly on top of each other for all visible $x$-values, your $t$ and $k$ are consistent with the given identity. Once verified, add your $t$ and $k$ values to obtain $t + k$.

Write the product so you can distribute:

Now distribute $x$ and $-3$:

Combine the $x^2$ terms and the $x$ terms:

So $(x^2 + tx + 6)(x - 3)$ simplifies to

We are told this expression is equivalent to

For two polynomials in $x$ to be equal for all $x$, the coefficients of each power of $x$ must match. So:

• Coefficient of $x^2$: $t - 3 = 2$
• Coefficient of $x$: $6 - 3t = k$

Solve the first equation for $t$:

Then substitute $t = 5$ into $6 - 3t = k$ to get $k$:

From the previous step,

So we have $t = 5$ and $k = -9$. Now add them:

Therefore, $t + k = -4$.