What are the solutions to the equation 4x-x^2-3=(x-2)(x^2-4x+3) ?

The solutions to the equation 4x-x^2-3=(x-2)(x^2-4x+3) are x=1,x=3

Find the zeros of 4x-x^2-3=(x-2)(x^2-4x+3)

The zeros of 4x-x^2-3=(x-2)(x^2-4x+3) are x=1,x=3

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

x = 1, x = 3

Solution steps

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

Expand

(x - 2) (x^{2} - 4 x + 3) : x^{3} - 6 x^{2} + 11 x - 6

4 x - x^{2} - 3 = x^{3} - 6 x^{2} + 11 x - 6

Switch sides

x^{3} - 6 x^{2} + 11 x - 6 = 4 x - x^{2} - 3

Move

3

to the left side

x^{3} - 6 x^{2} + 11 x - 3 = 4 x - x^{2}

Move

x^{2}

to the left side

x^{3} - 5 x^{2} + 11 x - 3 = 4 x

Move

4 x

to the left side

x^{3} - 5 x^{2} + 7 x - 3 = 0

Factor

x^{3} - 5 x^{2} + 7 x - 3 : (x - 1)^{2} (x - 3)

(x - 1)^{2} (x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x - 3 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 3 = 0 : x = 3

The solutions are

x = 1, x = 3

3 x^{4} - 10 x^{3} - 24 x^{2} - 6 x + 5 = 0

4 z^{5} - 14 z^{2} = 0

x^{3} + 5 x + 2 = 0

81 = r^{4}

x, 3 x^{3} + y^{4} = cy

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