Check if polynom is linear combination of given polynoms 
Given the polynomias  $$q_1(x)=2x+1 \\ q_2(x)=4x+1\\
 q_3(x)=x^3+x^2+x+1\\ q_4(x)=x^3+x-2  $$
is it possible to write $q_5(x)=2x^2+3x+4$ as a linear combination
  of $q_1,q_2,q_3,q_4$?

How do I tackle this?
 A: 
While writing the answer I stumbled upon a part in my lecture notes that answered my question. Maybe this can be of help to others.

You need to prove, that 
$$2x^2+3x+4 = a\cdot q_1 + b \cdot q_2 + c \cdot q_3 + d \cdot q_4$$
To do this, factorise the expression on the right side with the polynomials filled in:
$$2x^2+3x+4 = x^3(c+d)+x^2(c)+x(2a+4b+c+d)+a+b+c-2d$$
Now we get a linear system of equations:
$$0=c+d\tag{1}$$
 $$2=c\tag{2}$$
 $$3=2a+4b+c+d\tag{3}$$
 $$4=a+b+c-2d\tag{4}$$
By solving this, you will either get the scalars for a linear combination of the given polynomials to represent $q_5$ or evidence, that there's no such combination.
A: Hint: $$1=2q_1-q_2$$ $$x=\frac{1}{2}q_2-\frac{1}{2}q_1$$ $$x^2=q_3-q_4-3=q_3-q_4+3q_2-6q_1$$
A: Well, there's an easier (imho) way. Let's see if we can produce a canonical basis in $P_3[x]$ by using $q_{1..4}$.
Linear combinations of $q_1$ and $q_2$ allow us to generate polynomials $x$ and $1$, thus we can replace $q_{1..4}$ by polynomials $1$, $x$, $x^3+x^2$, $x^3$. Clearly, linear combinations of two last polynomials generate $x^2$ and $x^3$, thus $q_{1..4}$ are linearly independent and are the basis in $P_3[x]$ (because there're 4 of them and they generate $1$, $x$, $x^2$, $x^3$), hence $q_5$ is indeed a linear combination of $q_{1..4}$.
A: If we are looking at a linear combination with integer coefficients, let's check the parity of the coefficient of $x$.
In the first two polynomials this is even.
In the third and fourth the coefficient of $x$ is the same as the coefficient of $x^3$. To eliminate $x^3$, therefore, we have to reduce the contribution to the coefficient of $x$ from these two polynomials to zero too.
Therefore any $\mathbb Z-$linear combination of our given basis polynomials which has zero coefficient for $x^3$ has an even coefficient for $x$. Since this is not the case for the target polynomial, we can't make it.
Note that $($coefficient of $x^3+$ coefficient of $x)$ is always even, which is a more general parity statement.
