What are the solutions to the equation 2x^4-7x^3-6x^2+26x-12=0 ?

The solutions to the equation 2x^4-7x^3-6x^2+26x-12=0 are x=-2,x= 3/2 ,x=2+sqrt(2),x=2-sqrt(2)

Find the zeros of 2x^4-7x^3-6x^2+26x-12=0

The zeros of 2x^4-7x^3-6x^2+26x-12=0 are x=-2,x= 3/2 ,x=2+sqrt(2),x=2-sqrt(2)

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2 x^{4} - 7 x^{3} - 6 x^{2} + 26 x - 12 = 0

x = - 2, x = \frac{3}{2}, x = 2 + 2, x = 2 - 2

Decimal Notation

x = - 2, x = 1.5, x = 3.41421 \dots, x = 0.58578 \dots

Solution steps

2 x^{4} - 7 x^{3} - 6 x^{2} + 26 x - 12 = 0

Factor

2 x^{4} - 7 x^{3} - 6 x^{2} + 26 x - 12 : (x + 2) (2 x - 3) (x^{2} - 4 x + 2)

(x + 2) (2 x - 3) (x^{2} - 4 x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

2 x - 3 = 0 : x = \frac{3}{2}

Solve

x^{2} - 4 x + 2 = 0 : x = 2 + 2, x = 2 - 2

The solutions are

x = - 2, x = \frac{3}{2}, x = 2 + 2, x = 2 - 2

16 n^{9} - 25 n^{5} = 0

a^{20} = a 2

x^{2} (x - 4) (2 x + 5) = 0

10 = \frac{π}{2} D^{3}

5 x^{(4)} - 3 x^{(3)} + 4 x^{(2)} + 4 = 0

What are the solutions to the equation 2x^4-7x^3-6x^2+26x-12=0 ?
The solutions to the equation 2x^4-7x^3-6x^2+26x-12=0 are x=-2,x= 3/2 ,x=2+sqrt(2),x=2-sqrt(2)
Find the zeros of 2x^4-7x^3-6x^2+26x-12=0
The zeros of 2x^4-7x^3-6x^2+26x-12=0 are x=-2,x= 3/2 ,x=2+sqrt(2),x=2-sqrt(2)