Answers to Algebra Practical 6

Question 1 §

Factorise $p(x)$ into a product of irreducible factors in $\mathbb{Q}[x]$, showing that each of the factors is irreducible, such that

$$p(x)=x^3+4x^2+5x+6$$

$$\text{Using the rational root test, }\frac{r}{s}=\frac{\pm 6,\pm 3,\pm 2,\pm 1}{\pm 1}$$

$$\text{As all coefficients are positive, the only values of }\frac{r}{s}\text{ possible are negative:}$$

$$\frac{r}{s}=-1,-2,-3,-6$$

$$⟹\begin{array}{c}p(-1)=4\ne 0\\ p(-2)=4\ne 0\\ p(-3)=0\therefore \text{a rational root}\\ p(-6)=-6\ne 0\end{array}$$

$$\therefore p(x)=(x+3)(x^2+x+2)\text{this is a complete factorisation in }\mathbb{Q}[x].$$

Question 2 §

Find the complex roots of the biquadratic polynomial $p(x)$, and write its complete factorisations in $\mathbb{Q}[x]$, in $\mathbb{R}[x]$, and in $\mathbb{C}[x]$, such that

$$p(x)=3x^4-7x^2-6$$

$$⟹x^2=\frac{-(-7)\pm \sqrt{(-7{)}^2-4(3)(-6)}}{2(3)}\text{using the quadratic formula}$$

$$⟹p(x)=\left(x^2-3\right)\left(3x^2+2\right)\text{complete factorisation in }\mathbb{Q}[x].$$

$$⟹p(x)=(x+\sqrt{3})(x-\sqrt{3})(3x^2+2)\text{complete factorisation in }\mathbb{R}[x].$$

$$⟹p(x)=\frac{1}{3}(x+\sqrt{3})(x-\sqrt{3})(3x+i\sqrt{6})(3x-i\sqrt{6})\text{complete factorisation in }\mathbb{C}[x].$$

Question 3 §

Consider the self-reciprocal polynomial $p(x)=6x^4-35x^3+62x^2-35x+6$.

Question 3(i, ii, iii) §

Factorise the polynomial $p(x)$ into a product of three linear factors in $\mathbb{C}[x]$.

$$\text{Using the rational root test, }\frac{r}{s}=\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1,\pm 2,\pm 3,\pm 6}$$

$$\text{As it’s a self-reciprocal polynomial, we may exclude reciprocals such that }\frac{r}{s}=\frac{s}{r}.$$

$$\text{The polarity of the coefficients are alternating, thus we may exclude any }\frac{r}{s}<0.$$

$$⟹\frac{r}{s}=1,2,3,6,\frac{2}{3}$$

$$\text{Then, testing these potential roots:}\begin{array}{c}p(1)=4\\ p(2)=0\\ p(3)=0\\ p(6)=2244\\ p(\frac{2}{3})=\frac{28}{27}\end{array}$$

$$⟹x=2,3\text{ are valid roots}⟹x=\frac{1}{2},\frac{1}{3}\text{ are valid roots, because of the type of p(x)}$$

$$\therefore p(x)=(x-2)(x-3)(2x-1)(3x-1)\text{this is a complete factorisation in }\mathbb{C}[x].$$

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Correction in method:

$$6x^4-35x^3+62x^2-35x+6=0$$

$$⟹6\left(x^2+\frac{1}{x^2}\right)-35\left(x+\frac{1}{x}\right)+62=0$$

$$⟹6{\left(x+\frac{1}{x}\right)}^2-35\left(x+\frac{1}{x}\right)+50=0$$

$$\text{etc.}$$

Question 4 §

Factorise the following polynomial into a product of linear factors with complex coefficients, then factorise it into a product of irreducible factors over $\mathbb{R}$.

$$p(x)=x^4+(2-\sqrt{3})x^2-2\sqrt{3}$$

$$⟹x^2=a,b:ab=-2\sqrt{3},a+b=2-\sqrt{3}\therefore x^2=2,-\sqrt{3}$$

$$⟹p(x)=(x^2-2)(x^2+\sqrt{3})$$

$$⟹p(x)=(x-\sqrt{2})(x+\sqrt{2})(x-i\sqrt{3})(x+i\sqrt{3})$$