Factor. A) x^2-4x+3 B)x^2+5x-24

oldnick | Student

`B)`   `x^2+5x-24=0`

`Delta= 25-4(-24)=121>0`   Two real different solutions.


`x_1=3`    `x_2=-8`


oldnick | Student

`A)`    `x^2-4x+3`      find soluton of  `x^2-4x+3=0`

`Delta= 16-4(3)=4>0`  has two solution

`x=(4+-sqrt(4))/2=(4+-2)/2`     `x_1=3`     `x_2= 1`

The poliniom is decomponed in:

     `(x-x_1)(x-x_2)= (x-1)(x-3)`


spompeo4 | Student
A. (x-3)(x-1) b. (x+8)(x-3)
pramodpandey | Student

We have given

A.  `x^2-4x+3`

factorise 3 as 3.1




`` `=x(x-3)-1(x-3)`

`=(x-3)(x-1)`     ( brackets are equal so factor out )

`B. x^2+5x-24`

`we write 24=8xx3`

`x^2+(8-3)x-24`  ,

in bracket we take minus because last term has minus sign.



`=(x+8)(x-3)`   (brackets are same so factor out)

atyourservice | Student

 A) `x^2-4x+3`

use the a x c method

find factors of the product (3) that add up to b (-4) which would be -1 and -3 plug them in as b

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


`(x^2-x)(-3x+3) `

factor out he greatest common factor

x(x-1) -3(x-1)

(x-3) (x-1)

x=3  x=1


follow the same steps here

find factors of -24 that add up to b (5) which would be 8 and -3

plug them in as b

`x^2+8x-3x-24 `


`(x^2+8x)(-3x-24)  `

x(x+8) -3(x+8)


x=3   x=-8

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