What are the solutions to the equation 4x^3+4x^2-7x+2=0 ?

The solutions to the equation 4x^3+4x^2-7x+2=0 are x=-2,x= 1/2

Find the zeros of 4x^3+4x^2-7x+2=0

The zeros of 4x^3+4x^2-7x+2=0 are x=-2,x= 1/2

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4 x^{3} + 4 x^{2} - 7 x + 2 = 0

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

Solution steps

4 x^{3} + 4 x^{2} - 7 x + 2 = 0

Factor

4 x^{3} + 4 x^{2} - 7 x + 2 : (x + 2) (2 x - 1)^{2}

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

2 x - 1 = 0 : x = \frac{1}{2}

The solutions are

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

2 t^{4} + 4 t^{3} - 4 t^{2} = 0

16 x - 64 x^{3} = 0

x^{3} + 4 x^{2} + 6 x = 0

(x^{2} + 2 x) (x^{2} + 2 x - 7) = 8

- x^{4} + 7 x^{3} - 6 x^{2} = 0

What are the solutions to the equation 4x^3+4x^2-7x+2=0 ?
The solutions to the equation 4x^3+4x^2-7x+2=0 are x=-2,x= 1/2
Find the zeros of 4x^3+4x^2-7x+2=0
The zeros of 4x^3+4x^2-7x+2=0 are x=-2,x= 1/2