What are the solutions to the equation x^4+2x^3-43x^2-8x+336=0 ?

The solutions to the equation x^4+2x^3-43x^2-8x+336=0 are x=-3,x=4,x=-7

Find the zeros of x^4+2x^3-43x^2-8x+336=0

The zeros of x^4+2x^3-43x^2-8x+336=0 are x=-3,x=4,x=-7

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x^{4} + 2 x^{3} - 43 x^{2} - 8 x + 336 = 0

x = - 3, x = 4, x = - 7

Solution steps

x^{4} + 2 x^{3} - 43 x^{2} - 8 x + 336 = 0

Factor

x^{4} + 2 x^{3} - 43 x^{2} - 8 x + 336 : (x + 3) (x - 4)^{2} (x + 7)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 3 = 0 or x - 4 = 0 or x + 7 = 0

Solve

x + 3 = 0 : x = - 3

Solve

x - 4 = 0 : x = 4

Solve

x + 7 = 0 : x = - 7

The solutions are

x = - 3, x = 4, x = - 7

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

x^{3} + 13 x = 4 x^{2}

8 = a^{4}

(x^{2} - 7)^{2} + 4 (x^{2} - 7) - 5 = 0

- 2 x^{4} + 5 = 13

What are the solutions to the equation x^4+2x^3-43x^2-8x+336=0 ?
The solutions to the equation x^4+2x^3-43x^2-8x+336=0 are x=-3,x=4,x=-7
Find the zeros of x^4+2x^3-43x^2-8x+336=0
The zeros of x^4+2x^3-43x^2-8x+336=0 are x=-3,x=4,x=-7