What are the solutions to the equation x^2-16=(x^4)(x-4) ?

The solutions to the equation x^2-16=(x^4)(x-4) are x=4,x\approx-1.28378…,x\approx 1.53375…

Find the zeros of x^2-16=(x^4)(x-4)

The zeros of x^2-16=(x^4)(x-4) are x=4,x\approx-1.28378…,x\approx 1.53375…

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x^{2} - 16 = (x^{4}) (x - 4)

x = 4, x \approx - 1.28378 \dots, x \approx 1.53375 \dots

Solution steps

x^{2} - 16 = (x^{4}) (x - 4)

Expand

(x^{4}) (x - 4) : x^{5} - 4 x^{4}

x^{2} - 16 = x^{5} - 4 x^{4}

Switch sides

x^{5} - 4 x^{4} = x^{2} - 16

Move

16

to the left side

x^{5} - 4 x^{4} + 16 = x^{2}

Move

x^{2}

to the left side

x^{5} - 4 x^{4} + 16 - x^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

x^{5} - 4 x^{4} - x^{2} + 16 = 0

Factor

x^{5} - 4 x^{4} - x^{2} + 16 : (x - 4) (x^{4} - x - 4)

(x - 4) (x^{4} - x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 4 = 0 or x^{4} - x - 4 = 0

Solve

x - 4 = 0 : x = 4

Solve

x^{4} - x - 4 = 0 : x \approx - 1.28378 \dots, x \approx 1.53375 \dots

The solutions are

x = 4, x \approx - 1.28378 \dots, x \approx 1.53375 \dots

3 x^{3} - 15 x^{2} + 9 x - 5 = 0

108 x + 15 x^{2} - 3 x^{3} = 0

x, x^{3} + y^{2} - 4 y + 4 = 0

(x^{2} + 16 x)^{2} = 1 7^{2}

x^{3} - 3 x + 6 = 0

What are the solutions to the equation x^2-16=(x^4)(x-4) ?
The solutions to the equation x^2-16=(x^4)(x-4) are x=4,x\approx-1.28378…,x\approx 1.53375…
Find the zeros of x^2-16=(x^4)(x-4)
The zeros of x^2-16=(x^4)(x-4) are x=4,x\approx-1.28378…,x\approx 1.53375…