M1: Attempts to divide f(x) by ( x-2 ) and achieves a 3TQ quotient AND a remainder that is a constant.
Alternatively attempts to find the value of f(2). Allow f(2)=72 but if it is incorrect you must see some evidence for the award such as showing an intermediate line of 4×23+13×22−10×2+8
or else 32+52-20+8 (or equivalent) condoning slips. See equivalent form using synthetic division
A1: States that Remainder or R=72. It is implied if they write (ii) 72
It cannot just be awarded from sight of 72 within the division sum BUT if a candidate gets the division sum completely correct and does not state Q(x)=4x2+21x+32 and R=72 (o.e.) they should be awarded SC 1,1,1,0
Alt (a) via an identity BUT other acceptable methods exist such as by inspection which is a less formal version of this.
Sets up a correct identity f(x)=4x3+13x2−10x+8≡(x−2)(Ax2+Bx+C)+R AND then
M1: Compares terms and achieves A=4 and values for B and C
A1: States that Quotient, (i) or Q(x)=4x2+21x+32
M1: Finds R via f(2) or solving 8=R±2C with their value for C.
A1: States that Remainder, (ii) or R=72