Question

PTDTTNT:
\(\text{a) }6a^4+7a^3-37a^2-8a+12\)

\(\text{b) }\left(x^2+4x+8\right)^2+3x^3+14x^2+24x\)

 

Answer 1

a) \(6x^4+7x^3-37x^2-8x+12\)

\(=\left(6a^4+6a^3-36a^2\right)+\left(a^3+a^2-6a\right)+\left(-2a^2-2a+12\right)\)

\(=6a^2\left(a^2+a-6\right)+a\left(a^2+a-6\right)-2\left(a^2+a-6\right)\)

\(=\left(a^2+a-6\right)\left(6a^2+a-2\right)\)

Em làm tiếp nhé

b) Hướng dẫn:

=\(\left(x^2+4x+8\right)^2-\left(2x\right)^2+\left(2x\right)^2+3x^3+14x^2+24x\)

\(=\left(x^2+2x+8\right)\left(x^2+6x+8\right)+\left(3x^3+18x^2+24x\right)\)

\(=\left(x^2+6x+8\right)\left(x^2+2x+8+3x\right)\)

Em làm nhé!