I want to compute the following trace
$$Tr \Big( Y(\not{\!p_1'}+m) \Big) \ \ (1)$$
Where
$$\not{\!A} := \gamma^{\alpha} A_{\alpha} \ \ (2)$$
$$Y:= 4 \not{\!f_1} \not{\!p} \not{\!f_1} + m[-16(pf_1)+16 f_1^2] + m^2 ( 4 \not{\!p} - 16 \not{\!f_1})+16 m^3 \ \ (3)$$
The answer is given to be:
$$Tr \Big( Y(\not{\!p_1'}+m) \Big)=16 \Big( 2(f_1p)(f_1p') -f_1^2(pp')+m^2[-4(pf_1)+4f_1^2]+m^2 [(pp')-4(f_1 p')] +4m^4 \Big) \ \ (4)$$
We need to use the following properties
1) If $(\gamma^{\alpha}\gamma^{\beta}...\gamma^{\mu}\gamma^{\nu})$ contains an odd number of $\gamma$-matrices
$$Tr(\gamma^{\alpha}\gamma^{\beta}...\gamma^{\mu}\gamma^{\nu})=0$$
2)
$$Tr(\not{\!A}\not{\!B})=4AB$$
3)
$$Tr(\not{\!A}\not{\!B}\not{\!C}\not{\!D})=4\Big( (AB)(CD)-(AC)(BD)+(AD)(BC) \Big)$$
4) Given $A, B$ to be square matrices
$$Tr(A+B)=Tr(A) + Tr(B)$$
Besides: I've assumed that the trace of a scalar is equal to itself. (i.e. $Tr(m)=m$)
Applying such properties I get
$$Tr \Big( Y(\not{\!p_1'}+m) \Big) = 16 \Big( (f_1 p)(f_1 p') - f_1^2(pp') + f_1p'p p' \Big) + 16m^2p^2-64 m^2 f_1 p + 16m^2[-pf_1+f_1^2]+16m^4 \ \ (5)$$
Which is not the provided solution $(4)$. What am I missing?
Source: Second Edition QFT, Mandl & Shaw page 144.
Any help is appreciated.
The trace of a scalar is not the scalar itself. In this context, the scalar $m$ represents $m\mathbb1$, where $\mathbb1$ is the $4\times4$ identity matrix. Thus $\operatorname{Tr}m=4m$.