Question

Factorise
(a^2+3a-5)(a^2+3a+2)+6

Answer 1

Answer:

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Answer 2

The required factorised form is $a^4+6a^3+6a^2-9a+6$

Step-by-step explanation:

Since we have given that

$(a^2+3a-5)(a^2+3a+2)+6\\\\=a^2(a^2+3a+2)+3a(a^2+3a+2)-5(a^2+3a+2)+6\\\\=a^4+3a^3+2a^2+3a^3+9a^2+6a-5a^2-15a-10+6\\\\=a^4+6a^3+6a^2-9a+6$

Hence, the required factorised form is

$a^4+6a^3+6a^2-9a+6$

# learn more:

Factorise 5(3a+b)^2 + 6(3a+b) - 8

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