Polynomials Question (1 Viewer)

Ambility · Feb 3, 2016

How would you do this?

$P(x) is a monic polynomial of degree 4 with integer coefficients and constant term 3. One zero is i, another zero is rational and the sum of the zeros is negative. Factorise P(x) fully over \mathbb{R}.$

InteGrand · Feb 3, 2016

Ambility said:
How would you do this?

$P(x) is a monic polynomial of degree 4 with integer coefficients and constant term 3. One zero is i, another zero is rational and the sum of the zeros is negative. Factorise P(x) fully over \mathbb{R}.$

$\noindent Since the coefficients are real, another root is $-i$ by the conjugate root theorem. Note that there will be four roots in total. The constant term is 3, and $P$ is monic, so it is of the form $P(x) = x^4 + ax^3 +bx^2 + cx + 3$ (note that the coefficient of $x^3$ is negative as the sum of roots is negative), where $a,b,c \in \mathbb{Z}$. Now, since $P$ is an integer polynomial, any rational zero of $P$ must be either $\pm 3$ or $\pm 1$, by the rational roots theorem (factors of the constant term, since the polynomial is monic, so all rational roots are integers like this).$

$\noindent Since $i$ and $-i$ are roots, $\left(x^2 + 1\right)$ is a quadratic factor of $P(x)$. Let the other two roots be $r$ and $\beta$, where $r$ refers to the rational (in fact, integer) zero we were told $P$ possesses (so is one of $\pm 3, \pm1$). So $P(x) = \left(x^2 + 1\right)(x-r)(x-\beta)$. Note that $\beta$ must be real due to the conjugate root theorem (there aren't enough roots left for another complex conjugate pair of roots to exist).$

$\noindent Now, by expansion of the above, since $r$ is an integer, it is easy to see that $\beta$ must be rational (and hence an integer) for $P$ to have integer coefficients. (Alternatively, product of roots is 3, so $-i^2 \cdot r\beta = 3\Rightarrow r\beta = 3$. Whatever integer $r$ is from the possibilities for $r$, we see that $\beta$ will end up being another one from these so that the product is 3.) The only two possible integer roots of $P$ such that the sum of roots is negative and product be 3 is $-1,-3$, so these are values of $r$ and $\beta$. We have now found the roots, so we can factorise: $P(x) = \left(x^2 +1\right)(x+1)(x+3)$.$

Polynomials Question (1 Viewer)

Ambility

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InteGrand

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