Question

Phân tích đa thức thành nhân tử :

a, \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(b,\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+15\)

Answer 1

b, \(=\left(x-2\right)\left(x-8\right)\left(x-4\right)\left(x-6\right)+15\)

\(=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+15\)

Đặt \(x^2-10x+16\) là t , theo bài ra ta có:

\(=t\left(t+8\right)+15\)

\(=t^2+8t+16-1\)

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

\(=\left(t+3\right)\left(t+5\right)\)

Thay vào ta được:

\(=\left(x^2-10x+19\right)\left(x^2-10x+21\right)\)

CHÚC BẠN HỌC TỐT.....

Answer 2

a, Đặt \(x^2+x+1\) là a , theo bài ra ta có:

\(=t\left(t+1\right)-12\)

\(=t^2+t-12=\left(t^2+\dfrac{1}{2}.2.t+\left(\dfrac{1}{2}\right)^2\right)-12-\left(\dfrac{1}{2}\right)^2\)

\(=\left(t+\dfrac{1}{2}\right)^2-\dfrac{49}{4}\)

\(=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2\)

\(=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)\)

\(=\left(t-3\right)\left(t+4\right)\)

Thay vào ta được:

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

CHÚC BẠN HỌC TỐT..........

Answer 3

a) A = \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt \(x^2+x+1\) = a

A \(\Leftrightarrow a.\left(a+1\right)-12\) = \(a^2+a-12\) = \(\left(a-3\right)\left(a+4\right)\)

= \(\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)

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

b) B = \(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+15\)

B = \(\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+15\)

B = \(\left(x^2-10x+16\right)\left(x^2-10x+24\right)+15\)

Đặt \(x^2-10x+16\) = a

B \(\Leftrightarrow a.\left(a+8\right)+15\) = \(a^2+8a+15\)

= \(\left(a+3\right)\left(a+5\right)\)

= \(\left(x^2-10x+16+3\right)\left(x^2-10x+16+5\right)\)

= \(\left(x^2-10x+19\right)\left(x^2-10x+21\right)\)