Question

Hey guys

im new here
please solve this for me with steps!
ill mark as the best answer​

Answer 1

The factors of
`2(x+y)^2-9(x+y)-5` is ((x+y)-5)(2x+2y+1)

Explanation:

Given polynomial

=>
`2(x+y)^2-9(x+y)-5`

To Find:

The factors of the polynomial =?

Solution:

Lets assume k = (x+y)

Then
`2(x+y)^2-9(x+y)-5` can be written as
`2k^2-9k-5`

Now by using quadratic formula

k =
`(-b\pm√((b^2-4ac))/(2a)`

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =
`(-b\pm√((b^2-4ac)))/(2a)`

k =
`(-(-9) \pm √(((-9)^2-4(2)(-5)))/(2(2)))`

k =
`(-(-9) \pm √((81+40)))/(4)`

k =
`(-(-9) \pm √((121)))/(4)`

k =
`\frac{-(-9) \pm 11}}{4}`

k=
`( 9 \pm 11)/(4)`

k =
`(20)/(4)` k =
`(-2)/(4)`

`k_1 =5`
`k_2 = -(1)/(2)`

`2k^2-9k-5= 2(k-5)(k+(1)/(2))`

Solving the RHS we get

`(2)/(2)(k-5)(2k+1)`

`(k-5)(2k+1)`

Substituting k = x+y

`((x+y)-5)(2(x+y+1)`

`((x+y)-5)(2x+2y+1)`