Let $P(x)$ be a quadratic polynomial such that

$\left|\begin{array}{ll} P(0) & P(1) \\ P(0) & P(2) \end{array}\right|=0$

Let $P(0)=2$ and $P(1)+P(2)+P(3)=14$. Then $P(4)$ equals

Let's evaluate the determinant in the first condition:

$\left|\begin{array}{ll}P(0) & P(1) \\P(0) & P(2)\end{array}\right| = P(0)P(2) - P(0)P(1) = P(0)[P(2)-P(1)]$

Since this equals zero, and $P(0)=2 \neq 0$, we must have $P(2) = P(1)$.

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Since $P(x)$ is quadratic, we can write it as:

$P(x) = ax^2 + bx + c$

With $P(0) = c = 2$, our polynomial is:

$P(x) = ax^2 + bx + 2$

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Using $P(2) = P(1)$:

$P(1) = a + b + 2$

$P(2) = 4a + 2b + 2$

Setting them equal:

$a + b + 2 = 4a + 2b + 2$

$0 = 3a + b$

$b = -3a$

So our polynomial is now:

$P(x) = ax^2 - 3ax + 2$

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From $P(1) + P(2) + P(3) = 14$:

$P(1) = -2a + 2$

$P(2) = -2a + 2$

$P(3) = 9a - 9a + 2 = 2$

This gives us:

$(-2a + 2) + (-2a + 2) + 2 = 14$

$-4a + 6 = 14$

$-4a = 8$

$a = -2$

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With $a = -2$ and $b = -3a = 6$, our polynomial is:

$P(x) = -2x^2 + 6x + 2$

Therefore:

$P(4) = -2(16) + 6(4) + 2 = -32 + 24 + 2 = -6$