What are the solutions to the equation (x+6)(x+1)(x+3)(x-2)+56=0 ?

The solutions to the equation (x+6)(x+1)(x+3)(x-2)+56=0 are x=1,x=-5,x=2(sqrt(2)-1),x=-2(1+sqrt(2))

Find the zeros of (x+6)(x+1)(x+3)(x-2)+56=0

The zeros of (x+6)(x+1)(x+3)(x-2)+56=0 are x=1,x=-5,x=2(sqrt(2)-1),x=-2(1+sqrt(2))

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(x + 6) (x + 1) (x + 3) (x - 2) + 56 = 0

x = 1, x = - 5, x = 2 (2 - 1), x = - 2 (1 + 2)

Decimal Notation

x = 1, x = - 5, x = 0.82842 \dots, x = - 4.82842 \dots

Solution steps

(x + 6) (x + 1) (x + 3) (x - 2) + 56 = 0

Expand

(x + 6) (x + 1) (x + 3) (x - 2) + 56 : x^{4} + 8 x^{3} + 7 x^{2} - 36 x + 20

x^{4} + 8 x^{3} + 7 x^{2} - 36 x + 20 = 0

Factor

x^{4} + 8 x^{3} + 7 x^{2} - 36 x + 20 : (x - 1) (x + 5) (x^{2} + 4 x - 4)

(x - 1) (x + 5) (x^{2} + 4 x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 5 = 0 or x^{2} + 4 x - 4 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 5 = 0 : x = - 5

Solve

x^{2} + 4 x - 4 = 0 : x = 2 (2 - 1), x = - 2 (1 + 2)

The solutions are

x = 1, x = - 5, x = 2 (2 - 1), x = - 2 (1 + 2)

x^{5} = - 4 x^{5}

x^{3} + x = 100

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

130 r^{3} - a^{2} r^{3} + a^{2} = 0

x^{5} = 212

What are the solutions to the equation (x+6)(x+1)(x+3)(x-2)+56=0 ?
The solutions to the equation (x+6)(x+1)(x+3)(x-2)+56=0 are x=1,x=-5,x=2(sqrt(2)-1),x=-2(1+sqrt(2))
Find the zeros of (x+6)(x+1)(x+3)(x-2)+56=0
The zeros of (x+6)(x+1)(x+3)(x-2)+56=0 are x=1,x=-5,x=2(sqrt(2)-1),x=-2(1+sqrt(2))