Question

phân tích đa thức thành nhân tử

\(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

Answer 1

\(=2\left(x^2+x-5\right)^2-5\left(x^2+x-5\right)+3\)

\(=2\left(x^2+x-5\right)-2\left(x^2+x-5\right)-3\left(x^2+x-5\right)+3\)

\(=2\left(x^2+x-5\right)\left(x^2+x-6\right)-3\left(x^2+x-6\right)\)

\(=\left(x^2+x-6\right)\left(2x^2+2x-13\right)\)

\(=\left(x-2\right)\left(x+3\right)\left(2x^2+2x-13\right)\)

Answer 2

\(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

Đặt t=\(x^2+x\)

\(\Rightarrow C=2\left(t-5\right)^2-5t+28=2t^2-20t+50-5t+28=2t^2-25t+78=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)\)

Thay t: \(C=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)=2\left(x^2+x-\dfrac{13}{2}\right)\left(x^2+x-6\right)=2\left(x-2\right)\left(x+3\right)\left(x^2+x-\dfrac{13}{2}\right)\)

Answer 3

Ta có: \(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

\(=2\left(x^2+x-5\right)^2-5\left(x^2+x-5\right)+3\)

\(=\left(x-2\right)\left(x+3\right)\left(2x^2+2x-13\right)\)