$Q = \{18x^2 − 12x + 18, −36x^2 + sx + 9, 2sx + s\}$Find the two values of $s$ for which the set $Q$ is linearly dependent. Consider the subset of P3
$$Q  =  \{18x^2 − 12x + 18, −36x^2 + sx + 9, 2sx + s\}$$
Find the two values of s for which the set $Q$ is linearly dependent.
Does it mean that whenever we plug a value $s$ into $Q$, we will get a vector that is part of set $Q$?
And if that so, I can't seem to find two value $s$ that will make them linearly dependent, the first element $18x^2-12x+18$ is not affected by any value $s$, and to find two different value $s$ that make it linear dependent, I don't think it is possible
 A: hint
The third polynom does not contain the term with $ x^2 $. The only way to eliminate it from the first and the second, is multiplying the first by $ 2 $ and add it to the second. it gives
$$(36x^2-24x+36)+(-36x^2+sx+9)=$$
$$(s-24)x+45$$
this must be equal to
$$\lambda(2sx+s)$$
Now, solve the system
$$s-24=2\lambda.s$$
$$\lambda.s=45$$
A: We need to see can there exist scalars $\alpha, \beta, \gamma$ not all zero such that
\begin{align}
0 &= \alpha(18x^2-12x+18) + \beta(-36x^2+sx+9)+\gamma(2sx+s)\\
&= x^2(18\alpha - 36\beta) + x(-12\alpha + s\beta + 2s\gamma) + (18\alpha+9\beta+s\gamma)
\end{align}
or
$$\begin{cases}
18\alpha - 36\beta = 0 \\ -12\alpha+s\beta+2s\gamma = 0 \\ 18\alpha+9\beta+s\gamma = 0\end{cases}$$
The first equation implies $\alpha = 2\beta$ which we can plug into the second and third equation to obtain
$$\begin{cases}
(-24+s)\beta +2s\gamma = 0 \\ 45\beta + s\gamma = 0\end{cases}$$
Now multiplying the second equation by $-2$ and adding it to the first gives
$$(-114+s)\beta = 0.$$
If $s = 114$, we cannot conclude $\beta = 0$. In fact $\alpha = -57, \beta = -114, \gamma = 45$ is a nontrivial solution.
Moving on, if $s \ne 114$ we get $\beta = 0$ and then $s\gamma = 0$. If $s = 0$, we cannot conclude $\gamma = 0$. In fact $\alpha = \beta = 0, \gamma = 1$ is a nontrivial solution.
Moving on, if $s \ne 114, 0$ we get $\gamma = 0$ and then $\alpha = 0$ as well so there are no nontrivial solutions.
We conclude that $Q$ is linearly dependent if and only if $s \in \{0,114\}$.
