2x^4+3x^3-14x^2+8x

Question

What are all real zeros

Answer 1

2x⁴+3x³-14x²+8x=x(2x³+3x²-14x+8). x=0 is one of the zeroes.

The other zeroes are the zeroes of 2x³+3x²-14x+8, and they may be irrational.

One way of finding them is to use Newton's iterative Method which includes calculus.

If f(x)=2x³+3x²-14x+8, then f'(x)=6x²+6x-14.

x_n+1=x_n-f(x_n)/f'(x_n), where x₀ is close to a zero of f(x). A sketch of the graph of the cubic shows that there are zeroes near x=-4, x=1 and x=2. f(-4)=-128+48+56+8=-16; f(1)=-1; f(2)=16+12-28+8=8.

x₁=x₀-f(x₀)/f'(x₀).

x₀=-4:

x₁=-3.7241..., x₂=-3.6909..., x₃=-3.69047..., ..., x=-3.690471 after a few more iterations.

x₀=1:

x=0.755134 after a few iterations.

x₀=2:

x=1.435337 after a few iterations.

The zeroes (all real): -3.690471, 0, 0.755134, 1.435337 to 6 decimal place accuracy.