Given $f(x)+2 f(1-x)=x^{2}+3x.$ Replace $x \to 1-x$:

$

f(1-x)+2 f(x)=(1-x)^{2}+3(1-x)=x^{2}-5x+4.

$

Let $u=f(x),\ v=f(1-x)$. System: $u+2v=x^{2}+3x$, $2u+v=x^{2}-5x+4.$

$2\cdot\text{(second)} - \text{(first)}: 3u = 2x^{2}-10x+8 - (x^{2}+3x) = x^{2}-13x+8.$

So $f(x) = \dfrac{x^{2}-13x+8}{3}.$

$f(5) = \dfrac{25-65+8}{3} = \dfrac{-32}{3}.$ Hence $3 f(5) = -32.$

Verification: $f(5) + 2f(-4) = -32/3 + 2\cdot (16+52+8)/3 = -32/3 + 152/3 = 120/3 = 40 = 25 + 15.$ \checkmark

Why this is CAT-level: The natural false start -- guessing $f$ is a quadratic and matching coefficients -- works but is slower. The clean route uses the substitution $x\to 1-x$, but students must recognise that the involution $x \mapsto 1-x$ is the right swap (not $x \to -x$). Method selection under time pressure is the test.

Answer: $3 f(5) = -32.$