Let the quotient be $ax + b$ and write an equation:

Then expand the right-hand side.

After you expand, group terms by powers of $x$ and compare the coefficients of $x^3$, $x^2$, $x$, and the constant term. Use the simpler equations (like for $x^3$ and the constant) to find $a$ and $b$ first, then solve for $k$.

In Desmos, type f(x) = 4x^3 + 6x^2 - 14x + 7. Then type d(x) = x^2 + kx - 3, but replace k with the numerical value you found for $k$.

You already know from the algebra that the quotient should be $4x - 2$, so in Desmos type r(x) = f(x) - d(x)*(4x - 2). Also type g(x) = 2x + 1. If your value of $k$ is correct, the graphs of r(x) and g(x) will lie exactly on top of each other, confirming that the remainder is indeed $2x + 1$.

When you divide one polynomial by another, you can write

Here, the dividend is $4x^3 + 6x^2 - 14x + 7$, the divisor is $x^2 + kx - 3$, and the remainder is $2x + 1$.

Because a cubic divided by a quadratic has a linear quotient, let the quotient be $ax + b$.

So write the identity

Your goal is to find $a$, $b$, and $k$ so that this is true for all $x$. Then you will extract the value of $k$.

First expand $(x^2 + kx - 3)(ax + b)$:

Now add the remainder $2x + 1$:

So the identity from Step 1 becomes

Because these polynomials are equal for all $x$, their corresponding coefficients must match.

Match the coefficients of like powers of $x$ on both sides.

• Coefficient of $x^3$:

• Constant term:

so

Now you know the quotient is $4x - 2$. You still need to use the remaining coefficient equations to find $k$.

Use the $x^2$ coefficient equation:

• Coefficient of $x^2$:

Substitute $a = 4$ and $b = -2$:

Solve for $k$:

(You can also check with the $x$ coefficient equation $bk - 3a + 2 = -14$, which is satisfied by this same value of $k$.) So the value of $k$ is $2$.