# Thread: Need help with a few equations

1. ## Need help with a few equations

Hey,
surely those exercises are easy for you guys, but I need them to get done by thursday. I missed a few lessons, so I really don't know how to do these equations.

$\displaystyle x^3-3x^2+x+5=0$

$\displaystyle 2x^4-9x^3+4x^2+21x-18=0$ - Using the Horner scheme

Help would be appreciated.

2. Originally Posted by Rix12
Hey,
surely those exercises are easy for you guys, but I need them to get done by thursday. I missed a few lessons, so I really don't know how to do these equations.

$\displaystyle x^3-3x^2+x+5=0$

$\displaystyle 2x^4-9x^3+4x^2+21x-18=0$ - Using the Horner scheme

Help would be appreciated.
I don't know the Horner scheme so I shall do the first question

From inspection I know that $\displaystyle f(-1) = 0$

Thus using the factor theorem (see spoiler below for what it is) I know that (x+1) is a factor

Spoiler:
The factor theorem states that q(x) is a factor of f(x) if, and only if, f(x-q) =0

Which gives us $\displaystyle x^3-3x^2+x+5 = (x+1)(Ax^2+Bx+C)$ where A, B and C are constants to be found.

Comparing the coefficients:

• $\displaystyle x^3 \: \rightarrow \: 1 = A$
• $\displaystyle x^0 \: \rightarrow \: 5 = C$
• $\displaystyle x^1 \: \rightarrow \: 1 = B+C$

Thus $\displaystyle A=1$ , $\displaystyle B = -4$ and $\displaystyle C=5$

Put this back into our equation: $\displaystyle (x+1)(x^2-4x+5)$

Now we must check if the quadratic term will factor (ie if the discriminant is a square number or not): $\displaystyle \Delta = b^2-4ac = (-4)^2-4 \cdot 1 \cdot 5 = -4$.

This will not give a real solution so the final answer is $\displaystyle (x+1)(x^2-4x+5)=0$ thus $\displaystyle x=-1$ in the real numbers

Edit: Oops, forgot to see it was equal to 0 - complex solutions upcoming

$\displaystyle x^2-4x+5 = 0$

From the quadratic formula: $\displaystyle x = \dfrac{4 \pm \sqrt{(-4)^2-4 \cdot 1 \cdot 5}}{2} = \dfrac{4 \pm \sqrt{-4}}{2}$$\displaystyle \: = \dfrac{4\pm 2i}{2} = 2 \pm i$

where $\displaystyle i^2 = -1$

Thus all the solutions are: $\displaystyle x = -1, \: 2+i, \: 2-i$

3. Thank's a lot mate. Got the idea now.

Actually there's one more I forgot to ask:

Convert this one to algebraic form:
e^[(-1)-2*i]

4. Today I learned that Horner method not only computes the value of a polynomial $\displaystyle f(x)$ at a given point $\displaystyle x=a$ but also gives the ratio and the remainder of $\displaystyle f(x)$ and $\displaystyle x-a$.

Let $\displaystyle f_1(x)=2x^4-9x^3+4x^2+21x-18$. We can guess that $\displaystyle x_1=1$ is a root. (See below about how to make an educated guess.) Dividing $\displaystyle f_1(x)$ by $\displaystyle x-1$ gives $\displaystyle f_2(x)=2x^3-7x^2-3x+18$. (Instead of describing this division, I'll describe the next one.) Again by guessing, $\displaystyle x_2=2$ is a root of $\displaystyle f_2(x)$.

To divide $\displaystyle f_2(x)$ by $\displaystyle x-2$, make a table with three rows. The first row consists of the coefficients of $\displaystyle f_2$. The first element of the second row is 0. The third row is the sum of the first two. Therefore, the first element of the third row is 2.

Code:
 2 -7 -3  18
0  4 -6 -18
2 -3 -9   0
The second and each subsequent element of the second row is $\displaystyle x_2=2$ times the element of the third row immediately to the left. So, 4 = 2 * 2, -6 = 2 * (-3), and -18 = 2 * (-9). One computes the second and third row simultaneously left to right. The result indicates that the ratio is the polynomial $\displaystyle f_3(x)=2x^2-3x-9$ and the remainder is 0. Now, $\displaystyle f_3(x)=0$ is a quadratic equation.

To make an educated guess about the roots, one can use the rational root theorem. It says that if a rational number $\displaystyle p/q$ (in the lowest terms) is a root of $\displaystyle f(x)$, then $\displaystyle p$ divides the constant term of $\displaystyle f(x)$ and $\displaystyle q$ divides the leading coefficient. So, for $\displaystyle f_2(x)$, all rational roots are among $\displaystyle \pm1, \pm2, \pm 3, \pm 6, \pm 9, \pm 18, \pm 1/2, \pm 3/2, \pm 9/2$. Note that the theorem says nothing about irrational roots.

5. Originally Posted by Rix12
Thank's a lot mate. Got the idea now.

Actually there's one more I forgot to ask:

Convert this one to algebraic form:
e^[(-1)-2*i]
You are clearly expected to know two things in order to be able to answer this question:

1) $\displaystyle e^{a+ b}= e^{a}e^{b}$ and

2) $\displaystyle e^{ix}= cos(x)+ i sin(x)$.

6. Thanks guys. That helped a lot.