Question

Phần tích các đa thức sau thành nhân tử a.x^2+2x-3 b.x^2-10x+9 c.x^2-2x-15 d.x^2-2x-48

Answer 1

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

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

\(c,x^2-2x-15=x^2-5x+3x-15=\left(x^2-5x\right)+\left(3x-15\right)=\left(x-5\right)\left(x+3\right)\)

\(d,x^2-2x-48=x^2-8x+6x-48=\left(x^2-8x\right)+\left(6x-48\right)=\left(x-8\right)\left(x+6\right)\)

Answer 2

\(a,x^2+2x-3=x^2-x+3x-3=x\left(x-1\right)+3\left(x-1\right)=\left(x+3\right)\left(x-1\right)\\ b,x^2-10x+9=x^2-x-9x+9=x\left(x-1\right)-9\left(x-1\right)=\left(x-9\right)\left(x-1\right)\\ c,x^2-2x-15=x^2-5x+3x-15=x\left(x-5\right)+3\left(x-5\right)=\left(x+3\right)\left(x-5\right)\\ d,x^2-2x-48=x^2-8x+6x-48=x\left(x-8\right)+6\left(x-8\right)=\left(x+6\right)\left(x-8\right)\)

Answer 3

`a,`

`x^2 + 2x - 3`

`= x^2 - x + 3x - 3`

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

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

`b,`

`x^2 - 10x + 9`

`= x^2 - x - 9x + 9`

`= x ( x - 1 ) - 9 ( x - 1 )`

`= ( x - 9 ) ( x - 1 )`

`c)`

`x^2 - 2x - 15`

`= x^2 + 3x - 5x - 15`

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

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

`d)`

`x^2 - 2x - 48`

`= x^2 + 6x - 8x - 48`

`= x . ( x + 6 ) - 8 . ( x + 6 )`

`= ( x + 6 ) . ( x - 8 )`