How to write a polynomial factors given some of its roots when roots are not all > 1

Question

Given:

$$f(x)=3x^{2}+x-2\tag1$$ $$f(x) \ \textrm{has roots:} \ x = -1, \ x = \frac{2}{3}$$

My question: How to write $f(x)$ in factorized form, given the above information:

Option (A): $f(x)=(x+1)(x-2/3)$

OR

Option (B): $f(x)=(x+1)(3x-2)$

The correct answer is (B), but how can we decide to write the factor for corresponding to the root $x=2/3$ as $(3x-2)$ instead of $(x-2/3)$? In this case, it is easy to multiply the factored form and figure the correct answer, but what if the degree of the polynomial is high, how can we write the correct factorization given root value?

My first guess was that the rule could be:

if the root $r=\frac{p}{q}\:<1$ and neither $p=0$ nor $q=0$ and $a_n > 1$ (where $a_n$ is the coefficient of the highest root of p(x)), write the factor as $(qx-p)$ not as $(x-p/q)$. When $a_n=1$, write the factor as $(x-r)$ or as $(x-p/q)$

So, is the rule above correct? Do you know of any reference (I can reach via the internet) that says discuses such a case?

Thanks.

Answer 1

How you write the factors depends on the purpose of your factorisation. The factorisation is unique up to multiplication by units (a unit is a scalar with a multiplicative inverse).

So you could write:

$$3x^2+x-2=(3x-2)(x+1)=(-3x+2)(-x-1)=3\left(x-\frac 23\right)(x+1)$$

Note that to get the factorisation right, the last form has to have the scalar factor 3, otherwise the coefficient of $x^2$ is not right. Extracting the factor 3 is an allowed step if you are working over rational numbers (or real or complex numbers). However, if you want an integer factorisation, one of the first two is required.

Taking out the factor $3$ isolates the roots. The fact that this can't be done over the integers signals that there are non-integer roots.

If you are interested in the roots of the polynomial it is usual to take out the factors $-1$ and make the $x$ term in each case positive. But it can be convenient to do otherwise in the middle of a calculation.

Note also, for example, $$3x^2+xy-2y^2=(3x-2y)(x+y)$$ where the $x$ terms are positive, but the $y$ terms differ in sign. If you were more interested in $y$ than $x$ you might write instead $$2y^2-xy-3x^2=(2y-3x)(y+x)$$

Answer 2

A factorisation of $f$ over the integers, so in the ring $\Bbb Z[x]$, is option $B$ and not option $A$, because the factors then should be polynomials with integral coefficients. For a rational root, we obtain a factor $x-p/q$, and we always can rewrite the root $x-p/q=0$ as $qx-p=0$, so that we obtain a linear factor $qx-p$ with integral coefficients.

If the exercise says "factorize $3x^2+x-2$", then I would assume a factorisation in $\Bbb Z[x]$ with integer coefficients. By the Gauss Lemma, $f\in \Bbb Z[x]$ is irreducible if and only if it is irreducible and primitive in $\Bbb Q[x]$, see here.

Answer 3

'but what if the degree of the polynomial is high, how can we write the correct factorization given root value?'

You can't.

Let us assume your polynomial to be a quadratic. An infinite number of quadratic polynomials can be constructed that have the same two roots. This is because multiplying the polynomial with a constant value will result in another quadratic polynomial, but the roots will stay the same.

The example you mention is

$f(x)=3x^{2}+x-2\tag1$

Option A is just one of the polynomials having the roots -1 and 2/3. Here are some quadratic polynomials that have the same roots.

The polynomial in red is option A whereas the one in green is option B. Option b when expanded gives us the original polynomial (check this on your own). So you cannot write the correct factorization with just the root value. You also need to know the Coefficient of the term having the highest degree, which in this case is $x^2$.

The Coefficient of x in Option A is 1. But the coefficient of $x^2$ in f(x) is 3. So we multiply option A with 3 which gives us option B.

To generalise, If you have been given a polynomial of degree n wherein the coefficient of $x^n$ is a and the roots are say, $x_1, x_2, x_3, .....x_n$, then you can write the factorization as:

$a(x-x_1)(x-x_2)(x-x_3)..........(x-x_n)$

Answer 4

Option A: $$(x+1)(x-\frac 23)=x^2+\frac 13 x-\frac 23\ne 3x^3+x-2$$

If you expanded the factor form at option A, the resulting polynomial will have some coefficients rational numbers. It will be a different polynomial that only has the same roots as the original one.

Option B: $$(x+1)(3x-2)=3x^2+x-2$$ Therefore writing it in option B will render (by expanding) the polynomial in original form, with integer coefficients.

Notice that negating the factors in option B will still render the exact polynomial. That would not be a simplified factor form, that’s why is of no interest.

Conclusion: the simplest factor form that renders by expanding the original expression of the polynomial is the correct one.