Find a cubic polynomial to satisfy an integral equation

Find a cubic polynomial $P$ such that

$P(0) = P(-2) = 0, \qquad P(1) = 15, \qquad 3 \int_{-2}^0 P(x) \, dx = 4.$

Since $P(x)$ is a cubic polynomial we write,

$P(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3.$

Then,

$\begin{align*} P(0) &= 0 &\implies && c_0 &= 0 \\ P(-2) &= 0 &\implies && -c_1 + 2c_2 - 4c_3 &= 0 \\ P(1) &= 15 & \implies && c_1 + c_2 + c_3 &= 15 \end{align*}$

Finally, from the integral equation we have,

$\begin{align*} 3 \int_{-2}^0 P(x) \, dx = 4 && \implies && \int_{-2}^0 (c_1 x + c_2 x^2 + c_3 x^3) \, dx &= \frac{4}{3} \\ && \implies && \left. \left( \frac{c_1}{2} x^2 + \frac{c_2}{3} x^3 + \frac{c_3}{4} x^4 \right) \right|_{-2}^0 &= \frac{4}{3} \\ && \implies && -2c_1 + \frac{8}{3} c_2 - 4c_3 &= \frac{4}{3} \\ && \implies && -6c_1 + 8c_2 - 12c_3 &= 4 \\ && \implies && -3c_1 + 4c_2 - 6c_3 &= 2. \end{align*}$

Now, we have the following three equations in three unknowns,

$\begin{align*} \phantom{3}-c_1 + 2c_2 - 4c_3 &= 0 \\ \phantom{-3}c_1 + \phantom{2}c_2 + \phantom{4}c_3 &= 15 \\ -3c_1 + 4c_2 - 6c_3 &= 2. \end{align*}$

Since we don’t know any linear algebra, we’ll use elimination to solve for each variable. (If you know some linear algebra, you might know quicker ways to solve this system of equations.) First, adding the first and second equations we obtain

$3c_2 - 3c_3 = 15 \quad \implies \quad c_2 - c_3 = 5 \quad \implies \quad c_2 = 5+c_3.$