How do you solve a question like

Question

Suppose that a, b, and c are real numbers, and $P(x) = ax^2 + bx + c$. If $P(x + 2) + P(x + 1) − P(x) = 2x^2 + 15x + 20$, what is the value of $P(10)$?

This is a grade 8 Canadian Math Challenger's question (Regional, 2014 Round 1, Q24). Students get about 90 seconds to solve each question in this round. I tried different values of X but it seems that there has to be a trick to solve something like this in 90 or so seconds. This question can be found on page #525 of this PDF egbc.ca/getmedia/5cfa019a-f50b-4a12-97b0-e8ebf020ce71/

Answer 1

$$P(x) = ax^2 + bx + c$$ $$P(x+1) = a(x+1)^2 + b(x+1) + c=ax^2+(2a+b)x+(a+b+c)$$ $$P(x+2) = a(x+2)^2 + b(x+2) + c=ax^2+(4a+b)x+(4a+2b+c)$$ $$P(x+2)+P(x+1)-P(x)=ax^2+(6a+b)x+(5a+3b+c)=2x^2 +15x+20$$

Thus, comparing coefficients, we see that: $$a=2$$ $$12+b=15; b=3$$ $$10+9+c=20; c=1$$

Thus, $P(x)=2x^2+3x+1$

$P(10)=231$

Answer 2

Plugging numbers which have a P(0) somewhere in the equation may help you go a little faster to get your linear equations. Like so:

Plugging in $x=-1$:

$$2b+c = 7$$

Plugging in $x=0$:

$$5a+3b+c=20$$

Plugging in $x=-2$:

$$ -3a + b +c = -2$$

a=2, b=3, c=1

Then $P(10) = 231.$