Phân tích đa thức thành nhân tử :x3- x2 - 4
phân tích đa thức thành nhân tử
a x2 + 4x -y2 + 4
b 2x2 -18
c x3 -x2 -x + 1
d x2 -7xy + 10y2
a, \(x^2\) + 4\(x\) - y2 + 4
= (\(x^2\) + 4\(x\) + 4) - y2
= (\(x\) + 2)2 - y2
= (\(x\) + 2 - y)(\(x\) + 2 + y)
b, 2\(x^2\) - 18
= 2.(\(x^2\) -9)
= 2.(\(x\) -3).(\(x\) + 3)
c, \(x^3\) - \(x^2\) - \(x\) + 1
= (\(x^3\) + 1) - (\(x^2\) + \(x\))
= (\(x\) + 1)(\(x\)2 - \(x\) + 1) - \(x\).(\(x\) + 1)
=(\(x\) + 1).(\(x^2\) - \(x\) + 1 - \(x\))
= (\(x\) + 1).(\(x\) - 1)2
Phân tích đa thức thành nhân tử: x 3 + x 2 + y 3 + x y
A. ( x + y ) . ( x 2 - x y + y 2 + x )
B. ( x - y ) . ( x 2 + x y + y 2 - x )
C. ( x + y ) . ( x 2 + x y + y 2 - x )
D. ( x - y ) . ( x 2 + x y - y 2 + x )
phân tích đa thức thành nhân tử
x3 - x2 - x- 2
\(\Leftrightarrow x^3-2x^2+x^2-2x+x-2\)
\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)\)
`x^3-x^2-x-2`
`=x^3-2x^2+x^2-2x+x-2`
`=(x^3-2x^2)+(x^2-2x)+(x-2)`
`=x^2(x-2)+x(x-2)+(x-2)`
`=(x-2)(x^2+x+1)`
`@` `\text {Ans}`
`\downarrow`
`x^3 - x^2 - x - 2`
`= x^3 - 2x^2 + x^2 - 2x + x - 2`
`= (x^3 - 2x^2) + (x^2 - 2x) + (x-2)`
`= x^2(x - 2) + x(x - 2) + (x-2)`
`= (x^2 + x + 1)(x-2)`
x3-x2-x+1 → phân tích các đa thức thành nhân tử
\(x^2\left(x-1\right)-\left(x-1\right)=\left(x^2-1\right)\left(x-1\right)\)
\(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\)
\(x^3-x^2-x+1=\left(x^3-x^2\right)-\left(x-1\right)=x^2\left(x-1\right)-\left(x-1\right)=\left(x^2-1\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)^2\)
Phân tích đa thức sau thành nhân tử :
a,x4+8x+63
b,(x5+4)+(x3+4)-16
c,(x2+2x+7)+(x2-2x+4)(x2+2x+3)
a) \(x^4+8x+63\)
\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)
\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
c) \(\left(x^2+2x+7\right)+\left(x^2-2x+4\right)\left(x^2+2x+3\right)\left(1\right)\)
Ta có : \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Rightarrow x^2+2x+4=\dfrac{x^3-8}{x-2}\)
\(\left(1\right)\Rightarrow\left[\left(\dfrac{x^3-8}{x-2}+3\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-8}{x-2}-1\right)\right]\)
\(=\left[\left(\dfrac{x^3-3x-14}{x-2}\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-2x-5}{x-2}\right)\right]\)
\(=\dfrac{1}{x-2}\left[x^3-3x-14+\left(x^2-2x+4\right)\left(x^3-2x-5\right)\right]\)
phân tích đa thức thành nhân tử(2 nhóm hạng tử)
x3 -x2 +7x-7
\(x^3-x^2+7x-7=x^2\left(x-1\right)+7\left(x-1\right)=\left(x-1\right)\left(x^2+7\right)\)
Phân tích đa thức thành nhân tử:
a) x 4 + 1 - 2 x 2 ; b) x 2 - y 2 - 5y + 5x;
c) y 2 - 4 x 2 +4x - 1; d) x3 ( 2 + x ) 2 - ( x + 2 ) 2 + 1 - x 3 .
Phân tích các đa thức sau thành nhân tử:
1) x3 - 7x + 6
2) x3 - 9x2 + 6x + 16
3) x3 - 6x2 - x + 30
4) 2x3 - x2 + 5x + 3
5) 27x3 - 27x2 + 18x - 4
`1)x^3-7x+6`
`=x^3-x-6x+6`
`=x(x-1)(x+1)-6(x-1)`
`=(x-1)(x^2+x-6)`
`=(x-1)(x^2-2x+3x-6)`
`=(x-1)[x(x-2)+3(x-2)]`
`=(x-1)(x-2)(x+3)`
`2)x^3-9x^2+6x+16`
`=x^3-2x^2-7x^2+14x-8x+16`
`=x^2(x-2)-7x(x-2)-8(x-2)`
`=(x-2)(x^2-7x-8)`
`=(x-2)(x^2-8x+x-8)`
`=(x-2)[x(x-8)+x-8]`
`=(x-2)(x-8)(x+1)`
`3)x^3-6x^2-x+30`
`=x^3+2x^2-8x^2-16x+15x+30`
`=x^2(x+2)-8x(x+2)+15(x+2)`
`=(x+2)(x^2-8x+15)`
`=(x+2)(x^2-3x-5x+15)`
`=(x+2)[x(x-3)-5(x-3)]`
`=(x+2)(x-3)(x-5)`
`4)2x^3-x^2+5x+3`
`=2x^3+x^2-2x^2-x+6x+3`
`=x^2(2x+1)-x(2x+1)+3(2x+1)`
`=(2x+1)(x^2-x+3)`
`5)27x^3-27x^2+18x-4`
`=27x^3-9x^2-18x^2+6x+12x-4`
`=9x^2(3x-1)-6x(3x-1)+4(3x-1)`
`=(3x-1)(9x^2-6x+4)`
1) Ta có: \(x^3-7x+6\)
\(=x^3-x-6x+6\)
\(=x\left(x^2-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-6\right)\)
\(=\left(x-1\right)\left(x+3\right)\left(x-2\right)\)
2) Ta có: \(x^3-9x^2+6x+16\)
\(=x^3-2x^2-7x^2+14x-8x+16\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-7x-8\right)\)
\(=\left(x-2\right)\left(x-8\right)\left(x+1\right)\)
3) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
4) Ta có: \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+6\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+6\right)\)
5) Ta có: \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
Bài 2: Phân tích các đa thức sau thành nhân tử
a) x2 – 9 b) 4x2 -1 c) x4 - 16
d) x2 – 4x + 4 e) x3 – 8 f) x3 + 3x2 + 3x + 1
a) x² - 9
= x² - 3²
= (x - 3)(x + 3)
b) 4x² - 1
= (2x)² - 1²
= (2x - 1)(2x + 1)
c) x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x² - 2²)(x² + 4)
= (x - 2)(x + 2)(x + 4)
d) x² - 4x + 4
= x² - 2.x.2 + 2²
= (x - 2)²
e) x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)
f) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
Phân tích các đa thức sau thành nhân tử:
b ) x 3 – x 2 – 5 x + 125
b) x3 – x2 – 5x + 125
= (x3 + 125) - (x2 + 5x)
= (x + 5)(x2 - 5x + 25) - x(x + 5)
= (x + 5)(x2 - 5x + 25 - x)
= (x + 5)(x2 - 6x + 25)