Question

factorise: x2+9x+20+ax+4a​

Answer 1

Answer:

x2 + (9+a)x +(20+4a)=0

x2 + [(5+a)+4]x +(20+4a)=0

x2 + (5+a)x +4x + (20+4a)=0

x [x+(5+a)] + 4 [x+(5+a)]=0

(x+4)[x+(5+a)]=0

Step-by-step explanation:

here we will get two cases:-

1) when (x+4)=0

then x= -4

2) when [x+(5+a)]=0

then x= -(5+a)

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