Answer:

Here, p(z)=a(z)

3

+4(z)

2

+3z−4, q(z)=(z)

3

−4z+a, and the zero of z−3 is 3.

So, by the given condition

p(3)=q(3)

a(3)

3

+4(3)

2

+3(3)−4=(3)

3

−4(3)+a

27a+4×9+9−4=27−12+a

27a+36+9−4=27−12+a

27a+45−4=15+a

27a+41=15+a

27a−a=15−41

26a=−26

a=

26

−26

a=−1