\(3x^4+11x^3-7x^2-2x+1=\left(3x^4+12x^3-3x^2-3x\right)+\left(-x^3-4x^2+x+1\right)\)

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

a) \(4x^4+4x^3+5x^2+2x+1\)

= \(x^2\left(4x^2+4x+5+\frac{4}{x}+\frac{1}{x^2}\right)\)

=\(x^2\left[\left(4x^2+\frac{1}{x^2}\right)+2\left(2x+\frac{1}{x}\right)+5\right]\)(1)

Đặt \(2x+\frac{1}{x}=a\)thì \(\left(2x+\frac{1}{x}\right)^2=a^2\)\(\Rightarrow4x^2+\frac{1}{x^2}=a^2-4\)

Thay vào (1), ta có:

\(x^2\left(a^2-4+2a+5\right)\)

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

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

=\(\left[x\left(a+1\right)\right]^2\)

=\(\left[x\left(2x+\frac{1}{x}+1\right)\right]^2\)

=\(\left(2x^2+1+x\right)^2\)

\(=\left(2x^2+x+1\right)^2\)

a) Đặt f(x) = 4x⁴ + 4x³ + 5x² + 2x + 1

Sau khi phân tích thì đa thức có dạng ( 2x² + ax + 1 )( 2x² + bx + 1 )

=> f(x) = ( 2x² + ax + 1 )( 2x² + bx + 1 )

<=> f(x) = 4x⁴ + 2bx³ + 2x² + 2ax³ + abx² + ax + 2x² + bx + 1

<=> f(x) = 4x⁴ + ( a + b )2x³ + ( ab + 4 )x² + ( a + b )x + 1

Đồng nhất hệ số ta có : \(\hept{\begin{cases}a+b=2\\ab=1\end{cases}\Leftrightarrow}a=b=1\)

Vậy f(x) = 4x⁴ + 4x³ + 5x² + 2x + 1 = ( 2x² + x + 1 )²

b) 3x⁴ + 11x³ - 7x² - 2x + 1

= 3x⁴ - x³ + 12x³ - 4x² - 3x² + x - 3x + 1

= x³( 3x - 1 ) + 4x²( 3x - 1 ) - x( 3x - 1 ) - ( 3x - 1 )

= ( 3x - 1 )( x³ + 4x² - x - 1 )

somebody help me 

\(1,2x^2-3x-2\) 

\(=2x^2-4x+x-2\)

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

\(=\left(2x+1\right)\left(x-2\right)\) 

\(2,4x^2-7x-2\)

\(=4x^2-8x+x-2\) 

\(=4x\left(x-2\right)+x-2\)

\(\left(4x+1\right)\left(x-2\right)\)

\(=2x^4+6x^3-3x^3-9x^2-3x^2-9x+2x+6\)

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

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

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

2x^4+3x^3-12x^2-7x+6 = (2x^4-x^3)+(4x^3-2x^2)-(10x^2-5x)-(12x-6)

= x^3.(2x-1)+2x^2.(2x-1)-5x.(2x-1)-6.(2x-1) = (2x-1).(x^3+2x^2-5x-6) 

= (2x-1).[ (x^3+x^2)+(x^2+x)-(6x+6) ] = (2x-1).(x+1).(x^2+x-6) = (2x-1).(x-1).[(x^2-2x)+(3x-6)]

= (2x-1).(x+1).(x-2).(x+3)

k mk nha

a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)

\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)

\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)

\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)

\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)

\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)

\(=-\left(x-3\right)\left(x-2\right)\left(3x-8\right)\left(x+1\right)\)

b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)

\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)

\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)

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

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

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

bạn giúp mk 2 câu vừa đăng vs

2x³ + 3x² - 11x - 6 

Thử với x = 2 ta có :

2.2³ + 3.2² - 11.2 - 6 = 0

Vậy x = 2 là nghiệm của đa thức. Theo hệ quả của định lí Bézout thì đa thức trên chia hết cho x - 2

Thực hiện phép chia 2x³ + 3x² - 11x - 6 cho x - 2 ta được 2x² + 7x + 3

=> 2x³ + 3x² - 11x - 6 = ( x - 2 )( 2x² + 7x + 3 )

Lại có : 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x( x + 3 ) + ( x + 3 ) = ( x + 3 )( 2x + 1 )

=> 2x³ + 3x² - 11x - 6 = ( x - 2 )( x + 3 )( 2x + 1 )

Lại Bezout, thế này thì ...

1)  =\(-3x^4+9x^3+11x^3-33x^2-2x^2+6x-16x+48\)

    =\(-3x^3\left(x-3\right)+11x^2\left(x-3\right)-2x\left(x-3\right)-16\left(x-3\right)\)

=  \(\left(x-3\right)\left(-3x^3+11x^2-2x-16\right)\)

= \(\left(x-3\right)\left(-3x^3+6x^2+5x^2-10x+8x-16\right)\)

=\(\left(x-3\right)\left(-3x^2\left(x-2\right)+5x\left(x-2\right)+8\left(x-2\right)\right)\)

=  \(\left(x-3\right)\left(x-2\right)\left(-3x^2+5x+8\right)\)

= \(\left(x-3\right)\left(x-2\right)\left(x-\frac{8}{3}\right)\left(x+1\right)\)

Ý b lm theo ý tưởng tương tự nha bn :D