What are the solutions to the equation x^2(x-1)(2x+7)=0 ?

The solutions to the equation x^2(x-1)(2x+7)=0 are x=0,x=1,x=-7/2

Find the zeros of x^2(x-1)(2x+7)=0

The zeros of x^2(x-1)(2x+7)=0 are x=0,x=1,x=-7/2

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x^{2} (x - 1) (2 x + 7) = 0

x = 0, x = 1, x = - \frac{7}{2}

Decimal Notation

x = 0, x = 1, x = - 3.5

Solution steps

x^{2} (x - 1) (2 x + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or 2 x + 7 = 0

Solve

x - 1 = 0 : x = 1

Solve

2 x + 7 = 0 : x = - \frac{7}{2}

The solutions are

x = 0, x = 1, x = - \frac{7}{2}

16 x^{4} - 4 5^{3} + 9 x^{2} - 81 x + 81 = 0

0 = (x - 3) (x - 4) (x - 6)

2 = x^{3} - 7 x + 3

4 x^{3} - 4 x - 3 = 0

4 x^{4} - 169 x^{2} + 900 = 0

What are the solutions to the equation x^2(x-1)(2x+7)=0 ?
The solutions to the equation x^2(x-1)(2x+7)=0 are x=0,x=1,x=-7/2
Find the zeros of x^2(x-1)(2x+7)=0
The zeros of x^2(x-1)(2x+7)=0 are x=0,x=1,x=-7/2