What are the solutions to the equation 2(x-1)(x+2)=0 ?

The solutions to the equation 2(x-1)(x+2)=0 are x=1,x=-2

Find the zeros of 2(x-1)(x+2)=0

The zeros of 2(x-1)(x+2)=0 are x=1,x=-2

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2 (x - 1) (x + 2) = 0

x = 1, x = - 2

Solution steps

2 (x - 1) (x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

The solutions to the quadratic equation are:

x = 1, x = - 2

\frac{x ^{- 8} y ^{- 6} y ^{5}}{x ^{- 3}}

z^{4} + (\frac{i + 2 + i}{1 + i \cdot ( 2 + i )})^{3} \cdot (- i)

\frac{( 6 ) - 7}{( 6 ) ^{2} - 49}

125 x = 25

lo g_{8} (- 6 x) < 1

What are the solutions to the equation 2(x-1)(x+2)=0 ?
The solutions to the equation 2(x-1)(x+2)=0 are x=1,x=-2
Find the zeros of 2(x-1)(x+2)=0
The zeros of 2(x-1)(x+2)=0 are x=1,x=-2