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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.
- Using the Horner scheme
Help would be appreciated.
I don't know the Horner scheme so I shall do the first questionOriginally 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.
Help would be appreciated.
From inspection I know that
Thus using the factor theorem (see spoiler below for what it is) I know that (x+1) is a factor
Spoiler:
Which gives uswhere A, B and C are constants to be found.
Comparing the coefficients:
Thus,
and
Put this back into our equation:
Now we must check if the quadratic term will factor (ie if the discriminant is a square number or not):.
This will not give a real solution so the final answer isthus
in the real numbers
Edit: Oops, forgot to see it was equal to 0 - complex solutions upcoming
From the quadratic formula:^2-4 \cdot 1 \cdot 5}}{2} = \dfrac{4 \pm \sqrt{-4}}{2})
where
Thus all the solutions are:
Today I learned that Horner method not only computes the value of a polynomialat a given point
but also gives the ratio and the remainder of
.
Let. We can guess that
is a root. (See below about how to make an educated guess.) Dividing
by
gives
. (Instead of describing this division, I'll describe the next one.) Again by guessing,
is a root of
.
To divide, make a table with three rows. The first row consists of the coefficients of
. 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.
The second and each subsequent element of the second row isCode:2 -7 -3 18 0 4 -6 -18 2 -3 -9 0and the remainder is 0. Now,
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(in the lowest terms) is a root of
divides the constant term of
divides the leading coefficient. So, for
. Note that the theorem says nothing about irrational roots.
You are clearly expected to know two things in order to be able to answer this question: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]
1)and
2).
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