can you decompose a polynomial using an approximated zero?

Question

There's this exercise of numerical analysis in which I need to make a function that calculates the zeroes of a second grade polynomial and a function that applies newton's method, afterwards I need to calculate the solutions of a third grade polynomial, first finding one zero using the newton method and afterwards using the function for the solution of a second grade polynomial.

So I suppose that I should decompose the polynomial so that I can get a second grade equation to which apply the formula used to solve the quadratic equations, but can I really decompose a polynomial using an approximated solution?

Answer 1

You can but there may be some rounding errors.

For example, suppose you started with an arbitrary $$x^3 - 4 x^2 + x + 5=0$$ and found a solution using Newton's method of $x \approx -0.912$.

Then you could consider $$\frac{x^3 - 4 x^2 + x + 5}{x+0.912} = x^2-4.912 x+5.479744 +O(\tfrac1x)$$ i.e. $$x^3 - 4 x^2 + x + 5 \approx (x+0.912)(x^2-4.912 x+5.479744)$$ and solving $x^2-4.912 x+5.479744=0$ using the quadratic formula would suggest two more solutions of $x\approx 3.1991$ and $x\approx 1.7129$.

These are indeed close to the roots of the original equation, though $-0.912229$, $3.198691$ and $1.713538$ would be even closer.

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But to invent a difficulty, if the original equation had been $x^3-7x^2+16x-12=0$ with solutions $x=3,2,2$, and Newton's method had suggested $x \approx 3.00001$, then this approach could have suggested considering solving $\frac{x^3-7x^2+16x-12}{x-3.00001}\approx x^2-3.99999x+3.9999900001=0$ and that does not have real solutions though $x^2-4x+4=0$ does.

Answer 2

Yes, you can. Given an approximate root $a$, using polynomial division, you can write $p(x)=q(x)(x-a) + \epsilon$, and then given another approximate root $b$ of $q$, you have $p(x)=(r(x)(x-b)+\epsilon_2)(x-a)+\epsilon$, so $b$ is also an approximate root of $p$, as long as it isn't too far from $a$, with $p(b)=\epsilon+\epsilon_2(b-a)$.