Question

Factor each completely.one zero has been given

f(x)=x^3-4x+16;-2

F(x) = X^4+ 2x^-13x^2 -10x + 40;2

F(x)=x^3-3x^2-13x+15;-3

F(x)=x^4-8x^3+19x^2-12x;3

Answer 1

The polynomials in factored form are listed below:

1. f(x) = (x - 3.043) · (x - 1.521 - i 1.716) · (x - 1.521 + i 1.716)
2. f(x) = (x - 2.236) · (x - 2) · (x + 2.236) · (x + 4)
3. f(x) = (x + 3) · (x - 5) · (x - 1)
4. f(x) = (x - 4) · (x - 3) · (x - 1) · x

How to factor polynomials

In this question we must factor polynomials as there are both numerical and analytical methods. Mathematically speaking, factoring polynomials is represented by the following formula:

$\sum^{n}_{i = 0} \,c_{i}\cdot x^{i} = \prod^{n}_{i=1}(x-r_{i})$, $\forall\,x, r_{i}\in \mathbb{C}$   (1)

Where:

• $c_{i}$ - $i$-th Coefficient
• $r_{i}$ - $i$-th Root

Regarding fourth order polynomials we can solve them by Ferrari's method and third order polynomials by Descartes' method. Then, the solutions of each polynomials are given below:

Polynomial 1 ($f(x) = x^{3}-4\cdot x +16$)

f(x) = (x - 4) · (x - 3) · (x - 1) · x

Polynomial 2 ($f(x) = x^{4}+2\cdot x^{3}-13\cdot x^{2}-10\cdot x + 40$)

f(x) = (x - 2.236) · (x - 2) · (x + 2.236) · (x + 4)

Polynomial 3 ($f(x) = x^{3}-3\cdot x^{2}-13\cdot x +15$)

f(x) = (x + 3) · (x - 5) · (x - 1)

Polynomial 4 ($f(x) = x^{4}-8\cdot x^{3}+19\cdot x^{2}-12\cdot x$)

f(x) = (x - 4) · (x - 3) · (x - 1) · x

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