Answer:
The factors of [tex]2(x+y)^2-9(x+y)-5[/tex] is ((x+y)-5)(2x+2y+1)
Step-by-step explanation:
Given polynomial
=>[tex]2(x+y)^2-9(x+y)-5[/tex]
To Find:
The factors of the polynomial =?
Solution:
Lets assume k = (x+y)
Then [tex]2(x+y)^2-9(x+y)-5[/tex] can be written as [tex]2k^2-9k-5[/tex]
Now by using quadratic formula
k =[tex]\frac{-b\pm\sqrt{(b^2-4ac}}{2a}[/tex]
where
a= 2
b= -9
c= -5
Substituting the values, we get
k =[tex]\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}[/tex]
k =[tex]\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}[/tex]
k =[tex]\frac{-(-9) \pm \sqrt{(81+40)}}{4}[/tex]
k =[tex]\frac{-(-9) \pm \sqrt{(121)}}{4}[/tex]
k =[tex]\frac{-(-9) \pm 11}}{4}[/tex]
k= [tex]\frac{ 9 \pm 11}{4}[/tex]
k = [tex]\frac{20}{4}[/tex] k = [tex]\frac{-2}{4}[/tex]
[tex]k_1 =5[/tex] [tex]k_2 = -\frac{1}{2}[/tex]
[tex]2k^2-9k-5= 2(k-5)(k+\frac{1}{2})[/tex]
Solving the RHS we get
[tex]\frac{2}{2}(k-5)(2k+1)[/tex]
[tex](k-5)(2k+1)[/tex]
Substituting k = x+y
[tex]((x+y)-5)(2(x+y+1)[/tex]
[tex]((x+y)-5)(2x+2y+1)[/tex]
