Ta có : 5x² + 14x - 3

= 5x² + 15x - x - 3

= (5x² + 15x) - (x + 3)

= 5x(x + 3) - (x + 3)

= (x + 3)(5x - 1)

Phân tích đa thức thành nhân tử 5x2+14x−3

Theo đề bài ta có:

\(5x^2+14x-3\)

\(\Leftrightarrow5x^2+15x-x-3\)

\(\Leftrightarrow\left(5x^2+15x\right)-\left(x-3\right)\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x+3\right)\)

\(\Leftrightarrow\left(3x\right)\left(5x-1\right)\)

a, \(x^4+6x^3+7x^2-6x+1\)

\(=x^4-2x^2+1+6x^3+9x^2+6x\)

\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)

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

b, \(x^4-7x^3+14x^2-7x+1\)

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

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

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

c, \(12x^2-11x-36\)

\(=12x^2-27x+16x-36\)

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

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

a) \(x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1\)
Nhóm các hạng tử:

\(\left(\right. x^{5} - x^{4} \left.\right) + \left(\right. - 2 x^{3} + 2 x^{2} \left.\right) + \left(\right. x - 1 \left.\right) = \left(\right. x - 1 \left.\right) \left(\right. x^{4} - 2 x^{2} + 1 \left.\right) .\)

Đặt \(t = x^{2}\) thì \(x^{4} - 2 x^{2} + 1 = \left(\right. t - 1 \left.\right)^{2} = \left(\right. x^{2} - 1 \left.\right)^{2}\).
Vậy

\(\boxed{x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1 = \left(\right. x - 1 \left.\right) \left(\right. x^{2} - 1 \left.\right)^{2} = \left(\right. x - 1 \left.\right)^{3} \left(\right. x + 1 \left.\right)^{2} .}\)

b) \(x^{3} - 5 x^{2} - 14 x\)
Lấy \(x\) chung:

\(x \left(\right. x^{2} - 5 x - 14 \left.\right) = x \left(\right. x - 7 \left.\right) \left(\right. x + 2 \left.\right) .\)

\(\boxed{x^{3} - 5 x^{2} - 14 x = x \left(\right. x - 7 \left.\right) \left(\right. x + 2 \left.\right) .}\)

c) \(2 x^{2} + 2 x y - 4 y^{2}\)
Lấy \(2\) chung: \(2 \left(\right. x^{2} + x y - 2 y^{2} \left.\right)\).
Nhân tử hóa: \(x^{2} + x y - 2 y^{2} = \left(\right. x + 2 y \left.\right) \left(\right. x - y \left.\right)\).
\(\boxed{2 x^{2} + 2 x y - 4 y^{2} = 2 \left(\right. x + 2 y \left.\right) \left(\right. x - y \left.\right) .}\)

d) \(3 x^{2} + 8 x y - 3 y^{2}\)
Thử phân tích:

\(3 x^{2} + 8 x y - 3 y^{2} = \left(\right. 3 x - y \left.\right) \left(\right. x + 3 y \left.\right) .\)

\(\boxed{3 x^{2} + 8 x y - 3 y^{2} = \left(\right. 3 x - y \left.\right) \left(\right. x + 3 y \left.\right) .}\)

e) \(x^{2} - x - x y - 2 y^{2} + 2 y\)
Gộp lại theo \(x\): \(x^{2} + x \left(\right. - 1 - y \left.\right) + \left(\right. - 2 y^{2} + 2 y \left.\right)\).
Định thức là một bình phương → nghiệm \(x = 2 y\) và \(x = 1 - y\).
Vậy

\(\boxed{x^{2} - x - x y - 2 y^{2} + 2 y = \left(\right. x - 2 y \left.\right) \left(\right. x + y - 1 \left.\right) .}\)

f) \(x^{2} + 2 y^{2} - 3 x y + x - 2 y\)
Xem như phương trình bậc hai theo \(x\): nghiệm \(x = 2 y\) và \(x = y - 1\).
Do đó

\(\boxed{x^{2} + 2 y^{2} - 3 x y + x - 2 y = \left(\right. x - 2 y \left.\right) \left(\right. x - y + 1 \left.\right) .}\)

a: \(x^5-x^4-2x^3+2x^2+x-1\)

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

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

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

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

b: \(x^3-5x^2-14x\)

\(=x\left(x^2-5x-14\right)\)

\(=x\left(x^2-7x+2x-14\right)\)

=x[x(x-7)+2(x-7)]

=x(x-7)(x+2)

c: \(2x^2+2xy-4y^2\)

\(=2\left(x^2+xy-2y^2\right)\)

\(=2\left(x^2+2xy-xy-2y^2\right)\)

=2[x(x+2y)-y(x+2y)]

=2(x+2y)(x-y)

d: \(3x^2+8xy-3y^2\)

\(=3x^2+9xy-xy-3y^2\)

=3x(x+3y)-y(x+3y)

=(x+3y)(3x-y)

e: \(x^2-x-xy-2y^2+2y\)

\(=\left(x^2-xy-2y^2\right)-\left(x-2y\right)\)

\(=\left(x^2-2xy+xy-2y^2\right)-\left(x-2y\right)\)

=x(x-2y)+y(x-2y)-(x-2y)

=(x-2y)(x+y-1)

f: \(x^2+2y^2-3xy+x-2y\)

\(=x^2-2xy-xy+2y^2+x-2y\)

=x(x-2y)-y(x-2y)+(x-2y)

=(x-2y)(x-y+1)

\(x^3-5x^2-14x\)

\(=x^3+2x^2-7x^2-14x\)

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

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

\(=x\left(x+2\right)\left(x-7\right)\)

\(x^3-7x-6\)

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

\(=x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-6\right)\)

\(=\left(x+1\right)\left(x^2+2x-3x-6\right)\)

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

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

\(x^3-19x-30\)

\(=x^3-5x^2+5x^2-25x+6x-30\)

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

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

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

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

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

Đặt A = 12x² + 5x - 12y² + 12y - 10xy - 3 

= -(12x² + 10xy + 3 -12x² - 5x - 12y)

12y² + 10xy + 3 - 12x² - 5x - 12y

= 18xy + 12y² - 6y - 12x² - 8xy + 4x - 9x - 6y + 3

= 6y (3x + 2y - 1) - 4 . x (3x + 2y - 1) - 3 . (3x + 2y - 1) 

= (6y - 4x - 3) . (3x + 2y - 1)

\(x^4-14x^2-7x+30=\left(x^4+x^3-3x^2\right)+\left(-x^3-x^2+3x\right)+\left(-10x^2-10x+30\right)\)

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

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

x^4-14*x^2-7*x+30=(x^2-x-10)*(x^2+x-3)

=x³(x+2)-13x²+12x-26x+24

=x³(x+2)-x(13x-12)-2(13x-12)

=x³(x+2)-(13x-12)(x+2)

=(x+2)(x³-x-12x+12)

(x+2)[(x²-1)-12(x-1)]

=(x+2)[x(x-1)(x+1)-12(x-1)]

=(x+2)(x-1)[x(x+1)-12]

=(x+2)(x-1)(x²+x-12)

=(x+2)(x-1)(x²-3x+4x-12)

=(x+2)(x-1)[x(x-3)+4(x+3)]

=(x+2)(x-1)(x-3)(x+4)

trong bài làm của mk có hàng k có dấu "=" chỗ đó có dâu"=" nha!

\(x^4-14x^2-7x+30\)

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

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

Nhớ chọn cho tớ nhe!

1) \(\left(3x^2-3y^2\right)-\left(12x-12y\right)\)

\(=3xy\left(x-y\right)-12\left(x-y\right)\)

\(=\left(3xy-12\right)\left(x-y\right)\)

2) \(4x^3+4xy^2+8x^2y-16x\)

\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)

\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)

Ta có : 3x² - 3y² - 12x + 12y 

= (3x² - 3y²) - (12x - 12y)

= 3(x² - y²) - 12(x - y)

= 3(x - y)(x + y) - 4.3.(x - y)

= 3(x - y)(x + y - 4)