Basis for the set of polynomials of at most degree $2 $which intersect the $x-$axis at $x = 1$

Question

Let $P_2(\mathbb{R})$ be the set of all polynomials of at most degree $2$.

Define $W = \{p \in P_2(\mathbb{R}): p(1) = 0\}$.

Find the dimension of $W$.

My thoughts: $W$, in other words, are all polynomials of at most degree $2$ which intersect the $x$-axis at $x = 1$.

We know that $dim(P_2(\mathbb{R})) = 3$. ($a, b$ and $c$)

I could not think of a solution except that the $dim$ must be $2$ since $c$ (given $p = ax^2 + bx + c$) has to be $= (-a -b)$, therefore you can only have the basis: $x^2 - 1$ and $x - 1$.

Answer 1

Note that

$$p(x)=ax^2+bx+c \implies p(1)=0 \iff a+b+c=0$$

thus $p\in W$ has the form

$$p(x)=ax^2+bx-(a+b)=a(x^2-1)+b(x-1)$$

then your conclusion is correct.

Answer 2

The basis that you provided is indeed a basis of $W$. In fact:

• $x^2-1$ and $x-1$ are linearly independent, since none of them is the product of the other one by a scalar;
• if $ax^2+bx+c\in W$, then $a+b+c=0$ and therefore$$ax^2+bx+c=a(x^2-1)+b(x-1).$$

It follows from this that, as you wrote, $\dim W=2$.

Answer 3

Alternatively $p(1) = 0 \iff p = (ax+b)(x-1) = a(x^2-x) + b(x-1)$ which gives you another basis for W of $x^2-x$ and $x-1$.