Evaluation of vertex function with Feynman Parameters

On page 191 of Peskin & Schroeder, they show that after using Feynman parameters $x, y, z$, the denominator of the integrand of the vertex function is:

$$D = k^2 + 2k(yq – zp) + yq^2 + zp^2 – (x+y)m^2 + iepsilon,$$

where $p$ is the incoming fermion momentum, $k$ its momentum after it emits a photon, $k’$ after it interacts with the external field with momentum $q = k’-k$.

My question is how to simplify this expression after setting $ell := k +yq – zp$ to obtain:

$$D = ell^2 + xyq^2 – (1-z)^2m^2 + iepsilon.$$

I got stuck trying:

$$begin{aligned}D – ell^2 &= q^2(y-y^2) + p^2(z-z^2) + 2yzqp – (x+y)m^2
&=q^2y(1-y)+p^2z(1-z)+2yzqp – (1-z)m^2
&=q^2y(x+z)+p^2z(y+x) + 2yzqp – (1-z)m^2
&=xyq^2 + yz(q^2+p^2 + 2qp) +p^2zx – (1-z)m^2
&=xyq^2 + yz(q+p)^2 +p^2zx – (1-z)m^2.end{aligned}$$

Is it possible to cancel the two middle terms in the last line and obtain a square on the $(1-z)$?