Polynomials
Integer-coefficient polynomial with given values
MJAT_TS1_P2
Grade 12
Question:
Let $P(x)$ be a polynomial with integer coefficients satisfying $P(1) = 5$ and $P(2) = 7$. The smallest possible positive value of $P(10)$ is
Step-by-Step Solution
Key Concept: Write $P(x) = Q(x)(x-1)(x-2) + ax + b$ for integer polynomial $Q(x)$. Using $P(1)=5$: $a+b=5$. Using $P(2)=7$: $2a+b=7$. So $a=2$, $b=3$. Thus $P(x) = Q(x)(x-1)(x-2) + 2x+3$.
$P(x) = Q(x)(x-1)(x-2)+2x+3$ where $Q(x)$ is any polynomial with integer coefficients. $P(10) = Q(10)\cdot 9\cdot 8 + 23 = 72\cdot Q(10) + 23$. Minimum positive value when $Q(10)=0$: $P(10) = \mathbf{23}$.
Correct Answer: 23