Q.If $A = \begin{bmatrix}1 & 2\\ 4 & 1\end{bmatrix}$, find $A^2 + 2A + 7I$.

We are given a quartic polynomial $P(x)=2x^4+ax^3+bx^2+cx+d$ and four values: $P(1)=4$, $P(2)=7$, $P(3)=12$, $P(4)=19$.

We need $P(5)$.

Concept and intuition

The values $4,7,12,19$ are not random — look at their differences:

$7-4=3$, $12-7=5$, $19-12=7$. The second differences are constant: $5-3=2$, $7-5=2$.

That means these four points lie on a quadratic function. But $P(x)$ is a quartic (degree 4). So $P(x)$ must be that quadratic plus something that is zero at $x=1,2,3,4$ — a multiple of $(x-1)(x-2)(x-3)(x-4)$. This lets us find $P(5)$ without solving for all coefficients.

Step-by-step solution

1. Find the quadratic that fits the given points

   Let $Q(x)$ be a quadratic such that $Q(1)=4$, $Q(2)=7$, $Q(3)=12$, $Q(4)=19$.

   Since second differences are constant ($2$), the leading coefficient of $Q$ is $\frac{2}{2!}=1$.

   So $Q(x)=x^2+px+q$.

   Using $Q(1)=4$: $1+p+q=4 \Rightarrow p+q=3$.

   Using $Q(2)=7$: $4+2p+q=7 \Rightarrow 2p+q=3$.

   Subtract: $(2p+q)-(p+q)=3-3 \Rightarrow p=0$, then $q=3$.

   Thus $Q(x)=x^2+3$.

2. Relate $P(x)$ to $Q(x)$

   Since $P$ and $Q$ agree at four points, their difference is a polynomial of degree at most 4 that has roots at $x=1,2,3,4$.

   Hence

for some constant $k$.

3. Determine $k$ using the leading coefficient $P(x)=2x^4+ax^3+bx^2+cx+d$ has leading term $2x^4$. $Q(x)=x^2+3$ has leading term $0x^4$. The product $(x-1)(x-2)(x-3)(x-4)$ expands to $x^4 -10x^3 + \dots$, so its leading term is $x^4$. …