What is the answer to 2r(s+1)^2=0,2r^2(s+1)=0 ?

The answer to 2r(s+1)^2=0,2r^2(s+1)=0 is r=0,s=-1

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2r(s+1)^2=0,2r^2(s+1)=0

2 r (s + 1)^{2} = 0, 2 r^{2} (s + 1) = 0

r = 0, s = - 1

Solution steps

2 r (s + 1)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r = 0 or s + 1 = 0

r = 0

Solve

s + 1 = 0 : s = - 1

The solutions are

r = 0, s = - 1

(x + 2) (y - 1)^{3} (2 x - 1) = 0

a ((1) e^{- 6 t} - (t) 6 e^{- 6 t}) = 0

x^{2} y + x y - k^{2} y = 0

t e^{t} (c - 2 x) = 0

∣ a ∣ b = 0

What is the answer to 2r(s+1)^2=0,2r^2(s+1)=0 ?
The answer to 2r(s+1)^2=0,2r^2(s+1)=0 is r=0,s=-1