Phân tích đa thức sau thành nhân tử
a) x^3 + 4x^2 + 5x + 6
b) x^3 - 3x^2 - 4x + 12
c) 3x^3 - 7x^2 + 17x - 5
d) 2x^4 + 7x^3 - 2x^2 - 13x + 6
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
10 tháng 9 2021
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: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)
\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)
\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)
\(=-\left(x-1\right)\left(2x^3+2x^2+7x^2+7x+3x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\cdot\left(x+3\right)\left(2x+1\right)\)
H
21 tháng 8 2021
Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
21 tháng 8 2021
1: \(x^2+2x-3=\left(x+3\right)\left(x-1\right)\)
2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)
4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)
6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)
26 tháng 6 2023
2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)
15 tháng 12 2021
\(a,=x\left(x-2\right)^2\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\\ c,=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(x-2\right)\left(x+2\right)\left(2x-1\right)\\ d,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ e,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-y-2\right)\left(x+y-2\right)\\ g,=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\\ h,=x^3-x-2x+2=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x-2\right)=\left(x-1\right)^2\left(x+2\right)\\ i,=3x^2+3x-10x-10=\left(x+1\right)\left(3x-10\right)\)
SK
2
KC
23 tháng 8 2017
\(4x^3-13x^2+9x-18=4x^3-12x^2-x^2+3x+6x-18\)
\(=4x^2.\left(x-3\right)-x\left(x-3\right)+3.\left(x-3\right)=\left(x-3\right)\left(4x^2-x+3\right)\)
3 tháng 9 2018
\(4x^3-13x^2+9x-18\)
\(=4x^3-12x^2-x^2+3x+6x-18\)
\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)
\(=\left(x-3\right)\left(4x^2-x+6\right)\)
CW
15 tháng 7 2016
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
\(b,x^3-3x^2-4x+12\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(c,3x^3-7x^2+17x-5\)
\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)
\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)
\(\text{d) 2x}^4- 7x^3 - 2x^2 + 13x + 6\)
\(\text{= (2x^4 + 2x^3) - (9x^3 + 9x^2) + (7x^2 + 7x) + (6x + 6)}\)
\(\text{= 2x^3(x + 1) - 9x^2(x + 1) + 7x(x + 1) + 6(x + 1)}\)
\(\text{= (x + 1)(2x^3 - 9x^2 + 7x + 6)}\)
\(\text{= (x + 1)(2x + 1)(x - 3)(x - 2)}\)