Let $DM_{\text{gm}}$ be the category of Voevodsky´s geometric motives. Let $p,q\in \mathbb{Z}$ be integers with $p<0$.
Is it true that $$\text{Hom}_{DM_{\text{gm}}}(M_{\text{gm}}(X),\mathbb{Z}(p)[q])=0,$$ where $M(X)$ is the motive of a smooth scheme $X$ over a field $k$ and $\mathbb{Z}(p)$ is the Tate motive?
Here's an easy way to see this (which is more or less an elaboration of Mikhail's answer):
Suppose $i > 0$ and $n\in \mathbb Z$. We have the Thom isomorphism $$ H^n(X, \mathbb Z(-i)) \cong H^{n+2i}_X(\mathbb A^i_X, \mathbb Z(0)). $$ Weight $0$ motivic cohomology is just Zariski cohomology, so $$ H^n(X,\mathbb Z(0)) = \begin{cases} \mathbb Z^{\pi_0(X)} & \text{if }n=0,\\ 0 & \text{otherwise.}\end{cases} $$ Now the long exact sequence for cohomology with support shows that $H^*_X(\mathbb A^i_X,\mathbb Z(0))=0$.
I will sketch a proof.
It suffices to prove that there are only zero morphisms from $M_{\text{gm}}(X)(1)$ into $\mathbb{Z}[q]$ for any smooth $X$ and $q\in \mathbb{Z}$. The latter statement easily follows from the fact that the $\mathbb{Z}$ is a birational sheaf (with transfers); see Lemma 2.3.2(b) of "Birational Motives, II: Triangulated Birational Motives" by Bruno Kahn and Ramdorai Sujatha.
