Question

phân tích đa thức thành nhân tử : (x+3)^4+(x+1)^4-16

Answer 1

Đặt \(A=\left(x+3\right)^4+\left(x+1\right)^4-16\)

\(=\left\lbrack\left(x+2\right)+1\right\rbrack^4+\left\lbrack\left(x+2\right)-1\right\rbrack^4-16\)

Đặt b=x+2

=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)

\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)

\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)

\(=2\left(b^2+1\right)^2+8b^2-16\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)

\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)

\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)

\(=2\left\lbrack\left(x+2\right)^2+7\right\rbrack\left(x+2-1\right)\left(x+2+1\right)=2\left(x+1\right)\left(x+3\right)\left\lbrack\left(x+2\right)^2+7\right\rbrack\)

Answer 2

Đặt \(A=\left(x+3\right)^4+\left(x+1\right)^4-16\)

\(=\left\lbrack\left(x+2\right)+1\right\rbrack^4+\left\lbrack\left(x+2\right)-1\right\rbrack^4-16\)

Đặt b=x+2

=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)

\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)

\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)

\(=2\left(b^2+1\right)^2+8b^2-16\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)

\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)

\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)

\(=2\left\lbrack\left(x+2\right)^2+7\right\rbrack\left(x+2-1\right)\left(x+2+1\right)=2\left(x+1\right)\left(x+3\right)\left\lbrack\left(x+2\right)^2+7\right\rbrack\)