Given

f(x)=ax2+bx+2 and
g(x)=bx2+ax+1
x−2 is a factor of f(x)

∴ By factor theorem,
f(2)=0
⇒a(2)2+b(2)+2=0
⇒4a+2b+2=0
By dividing both sides by 2,
⇒2a+b+1=0…(i)

Also given that, g(x) divide dby (x−2), leaves remainder −15
∴ By remainder theorem,
g(2)=−15
⇒b(2)2+2a+1=−15
⇒4b+2a+1+15=0
⇒4b+2a+16=0...(ii)

Now, subtracting equation (i) from equation (ii), we get
(4b+2a+16)–(2a+b+1)=0–0
⇒4b+2a+16−2a−b−1=0
⇒3b+15=0
⇒3b=−15
⇒b=−315​=−5

Substituting this value in equation (i), we get
2a+b+1=0
⇒2a−5+1=0
⇒2a−4=0
⇒2a=4
⇒a=2

∴f(x)=ax2+bx+2=2x2–5x+2
g(x)=bx2+ax+1=−5x2+2x+1

Then, f(x)+g(x)+4x2+7x
=2x2–5x+2–5x2+2x+1+4x2+7x
=x2+4x+3
=x2+3x+x+3
=x(x+3)+1(x+3)
=(x+1)(x+3)