Answers to Algebra Practical 5

Question 1 §

$$\frac{4x^3-3x+9}{2x^3+x^2-3x+6}$$

First, finding the greatest common divisor of both the numerator and the denominator,

$$(4x^3-3x+9)=(2x^3+x^2-3x+6)q(x)+r(x)$$

$$⟹(4x^3-3x+9)=(2x^3+x^2-3x+6)(2)-(2x^2-3x+3)$$

$$⟹(2x^3+x^2-3x+6)=(2x^2-3x+3)(x)+2(2x^2-3x+3)$$

$$⟹(2x^2-3x+3)=(2x^2-3x+3)(1)+(0)$$

$$\therefore \text{GCD}=2x^2-3x+3$$

Using this to then factorise the fraction,

$$⟹\frac{4x^3-3x+9}{2x^3+x^2-3x+6}=\frac{(2x^2-3x+3)(2x+3)}{(2x^2-3x+3)(x+2)}=\frac{2x+3}{x+2}$$

Question 2 §

$$(2x^2+x+3)\cdot u(x)+(x^2+1)\cdot v(x)=1$$

First, finding the greatest common divisor of both the numerator and the denominator,

$$(2x^2+x+3)=(x^2+1)(2)+(x+1)$$

$$⟹(x^2+1)=(x+1)(x-1)+2$$

(working for that division is here)

$$⟹(x+1)=2(\frac{1}{2}x+\frac{1}{2})+0$$

$$\therefore \text{GCD}=2,\text{ or any other integer, e.g. 1}:\text{coprime}$$

Next, using the extended Euclidean algorithm,

$$2=(x^2+1)-(x+1)(x-1)$$

$$⟹2=(x^2+1)-(x-1)((2x^2+x+3)-2(x^2+1))$$

$$=(x^2+1)-(x-1)(2x^2+x+3)+2(x-1)(x^2+1)$$

$$=(x^2+1)(2x-1)+(1-x)(2x^2+x+3)$$

$$\therefore (2x^2+x+3)(\frac{1}{2}-\frac{1}{2}x)+(x^2+1)(x-\frac{1}{2})=1$$

So, $u(x)=\frac{1}{2}-\frac{1}{2}x$ and $v(x)=x-\frac{1}{2}$. Note that I’m not sure if that’s the end of the question, or if we need to introduce $n$.

Question 3 §

$$(x^3+x+1)\cdot u(x)+(x^2-x-1)\cdot v(x)=1$$

First, finding the greatest common divisor of both the numerator and the denominator,

$$(x^3+x+1)=(x^2-x-1)(x+1)+(3x+2)$$

(working for that division is here)

$$⟹(x^2-x-1)=(3x+2)(\frac{3}{9}x-\frac{5}{9})-(\frac{1}{9})$$

(working for that division is here)

$$⟹9(x^2-x-1)=(3x+2)(3x-5)-(1)$$

$$⟹(3x+2)=(1)(3x+2)+(0)$$

$$\therefore \text{GCD}=1:\text{coprime}$$

Next, using the extended Euclidean algorithm,

$$1=9(x^2-x-1)-(3x+2)(3x-5)$$

$$⟹1=9(x^2-x-1)-((x^3+x+1)-(x+1)(x^2-x-1))(3x-5)$$

$$=9(x^2-x-1)-(3x-5)(x^3+x+1)+(3x-5)(x+1)(x^2-x-1)$$

$$=(9+(3x-5)(x+1))(x^2-x-1)-(3x-5)(x^3+x+1)$$

$$=(x^3+x+1)(5-3x)+(x^2-x-1)(3x^2-2x+4)=1$$

$$\therefore u(x)=5-3x,v(x)=3x^2-2x+4$$

Question 4 §

$$10^6-3^6=999271$$

$$⟹(10^3+3^3)(10^3-3^3)=999271$$

$$⟹(10+3)(10^2-(10)(3)+3^2)(10-3)(10^2+(10)(3)+3^2)=999271$$

$$⟹(13)(79)(7)(139)=999271$$

$$⟹999271=139\cdot 79\cdot 13\cdot 7:\text{these are prime}\square$$

Question 5 §

Question 5a §

$$f(x)=x^4+2x^3-6x^2+8x+80$$

(2+2i)	1	2	-6	8	80
	0	2+2i	4+12i	-28+20i	-80
	1	4+2i	-2+12i	-20+20i	0

$$⟹f(x)=[x-(2+2i)][x^3+(4+2i)x^2+(-2+12i)x+(-20+20i)]$$

Question 5b §

If $z$ is a root, then $\bar{z}$ will be a root, so using Ruffini’s Rule we can decompose the cubic as well:

(2-2i)	1	4+2i	-2+12i	-20+20i
	0	2-2i	12-12i	20-20i
	1	6	10	0

$$⟹f(x)=[x-(2+2i)][x-(2-2i)][x^2+6x+10]$$

Then we can use the quadratic formula for the remaining two roots:

$$x^2+6x+10:\frac{-b\pm \sqrt{b^2-4ac}}{2a}⟹\frac{-6\pm \sqrt{6^2-4\cdot 1\cdot 10}}{2}⟹-3\pm i$$

Thus the final factorised form of $f(x)$ is:

$$\therefore f(x)=[x-(2+2i)][x-(2-2i)][x-(-3+i)][x-(-3-i)]$$

$$\text{The roots are: }2+2i,2-2i,-3+i,-3-i.$$

Question 6 §

(1)	1	0	…	0	2	0	…	0	1
	0	1	1	1	1	3	3	3	3
	1	1	1	1	3	3	3	3	4

$$(x^{100}+2x^6+1)=(x-1)(x^{100}+x^{99}+x^{98}+\cdots +x^6+3x^5+3x^4+\cdots +3x+3)+4$$

Question 7 (may be incorrect) §

Using the following points,

$$f(-1)=-7,f(0)=-3,f(1)=-3,f(2)=-1$$

we can model a unique polynomial $f(x)$:

$$f(x)=a{x}^3+b{x}^2+cx+d$$

$$⟹-7=-a+b-c+d,-3=d,-3=a+b+c+d-1=8a+4b+2c+d$$

$$⟹-4=-a+b-c,6=a+b+c2=8a+4b+2c$$

$$⟹\left[\begin{array}{ccc}-1& 1& -1\\ 1& 1& 1\\ 8& 4& 2\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}-4\\ 6\\ 2\end{array}\right]$$

$$⟹\left[\begin{array}{c}a\\ b\\ c\end{array}\right]={\left[\begin{array}{ccc}-1& 1& -1\\ 1& 1& 1\\ 8& 4& 2\end{array}\right]}^{-1}\left[\begin{array}{c}-4\\ 6\\ 2\end{array}\right]$$

$$⟹\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}-2\\ 1\\ 7\end{array}\right]$$

$$\therefore f(x)=-2x^3+x^2+7x-3$$

Question 8 §

$$f(x)=a_n{x}^n+\cdots +a_{1}x+a_0$$

Question 8a §

Considering the sum of only the even order coefficients,

$$S_{even}=a_n+a_{n-2}+a_{n-4}+\cdots +a_4+a_2+a_0$$

We can then consider the behaviour of $f(x)$ at $x=\pm 1$,

$$f(1)=a_n+a_{n-1}+a_{n-2}+\cdots +a_2+a_1+a_0$$

$$f(-1)=a_n-a_{n-1}+a_{n-2}-\cdots +a_2-a_1+a_0$$

where we see that $f(1)$ is the sum of all coefficients, and $f(-1)$ is an alternating sequence of adding and subtracting the coefficients - adding the even terms, subtracting the odd.

Thus, if we find the sum of the two sequences, we’ll end with $2S_{even}$, where ${}_{even}S$ was defined earlier:

$$f(1)+f(-1)=2a_n+2a_{n-2}+2a_{n-4}+\cdots +2a_4+2a_2+2a_0=2S_{even}$$

$$⟹S_{even}=\frac{1}{2}(f(1)+f(-1))\square$$

Question 8b §

Considering the sum of only the odd order coefficients,

$$S_{odd}=a_{n-1}+a_{n-3}+a_{n-5}+\cdots +a_3+a_1$$

We can then consider the behaviour of $f(x)$ at $x=\pm 1$,

$$f(1)=a_n+a_{n-1}+a_{n-2}+\cdots +a_2+a_1+a_0$$

$$f(-1)=a_n-a_{n-1}+a_{n-2}-\cdots +a_2-a_1+a_0$$

where we see that $f(1)$ is the sum of all coefficients, and $f(-1)$ is an alternating sequence of adding and subtracting the coefficients - adding the even terms, subtracting the odd.

Thus, if we find the difference of the two sequences, we’ll end with $2S_{odd}$, where $S_{odd}$ was defined earlier:

$$f(1)-f(-1)=2a_{n-1}+2a_{n-3}+2a_{n-5}+\cdots +2a_5+2a_3+2a_1=2S_{odd}$$

$$⟹S_{odd}=\frac{1}{2}(f(1)-f(-1))\square$$

Question 9 §

If we want to find the sum of the coefficients of a polynomial, we can substitute $x=1$ and have that returned to us - this is irrelevant of what form it is in.

So, if we consider the polynomial

$$p(x)=(2x-1)(3x^2-2)(4x^3-3)(5x^4-4)(6x^5-5)$$

then the sum of its coefficients would be

$$p(1)=(1)(1)(1)(1)(1)=1\square$$

Note that I initially tried to solve this using sequences and combinations, which may have eventually worked, but then I realised I could just do this.