Question

Hey guysim new hereplease solve this for me with steps!ill mark as the best answer​

Answer 1

Answer:

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

Step-by-step 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 =$\frac{-b\pm\sqrt{(b^2-4ac}}{2a}$

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =$\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}$

k =$\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}$

k =$\frac{-(-9) \pm \sqrt{(81+40)}}{4}$

k =$\frac{-(-9) \pm \sqrt{(121)}}{4}$

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

k= $\frac{ 9 \pm 11}{4}$

k =  $\frac{20}{4}$                         k =  $\frac{-2}{4}$    

$k_1 =5$                                      $k_2 = -\frac{1}{2}$

$2k^2-9k-5= 2(k-5)(k+\frac{1}{2})$

Solving the RHS we get

$\frac{2}{2}(k-5)(2k+1)$

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

Substituting k = x+y

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

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