Simultaneous polynominals

  • Thread starter ghostbuster25
  • Start date
So if you had a root of -1, then (x+1) is a factor and when you evaluate f(-1) you will get 0. This is the same concept, but instead of 0, the polynomial is divisible by (x+1).In summary, to find the values of a and b for the cubic polynomial f(x)= 4x^3+ax^2+bx+6, which has a factor (x-2) and when divided by (x+1) has a remainder of -15, evaluate f(-1) and set it equal to -15 to solve for a and b. The correct way to evaluate f(-2) is to set it equal to 0 since
  • #1
ghostbuster25
102
0
simultaneous polynominals!

stuck on this coursework question
The cubic polynominal f(x)= 4x^3+ax^2+bx+6 has a factor (x-2)
when it is divided by (x+1) it has a remainder 0f -15

find the values of a and b

I did as follows
factor (x-2)
f(-2)
4(-2)^3+a(-2)^2+b(-2)+6=0
equals 4a-2b=26

when it is divided by (x+1)
f(-1)
4(-1)^3+a(-1)^2+b(-1)+6=15
equals a-b=-17

when i solve these it gives me b=17 and a=-8.5

Which i know is wrong

plese tell me where

thanks
 
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  • #2


ghostbuster25 said:
stuck on this coursework question
The cubic polynominal f(x)= 4x^3+ax^2+bx+6 has a factor (x-2)
when it is divided by (x+1) it has a remainder 0f -15

find the values of a and b

I did as follows
factor (x-2)
f(-2)
4(-2)^3+a(-2)^2+b(-2)+6=0
equals 4a-2b=26

when it is divided by (x+1)
f(-1)
4(-1)^3+a(-1)^2+b(-1)+6=15
equals a-b=-17

when i solve these it gives me b=17 and a=-8.5

Which i know is wrong

plese tell me where

thanks

You mistakenly found f(-2) when dividing by (x-2)
 
  • #3


i thought i was just to put the factor into the equation. Not sure what other way to do. Is the second part correct?
 
  • #4


First, divide [itex](x-2)[/tex] into [itex]4x^3+ax^2+bx+6[/tex]
The remainder will have a- and b-terms in it.
But, since [itex](x-2)[/tex] is a factor, that means that it divides evenly into [itex]4x^3+ax^2+bx+6[/tex], so the remainder must be 0.

In other words, you'll get a remainder of the form ad + be + f (where d, e, and f are integers). Set this term equal to 0.

Now, divide [itex](x+1)[/tex] into [itex]4x^3+ax^2+bx+6[/tex].
Again, you'll get a remainder with a- and b-terms in it.
Set this remainder to -15.

You now have 2 equations with 2 variables (a and b) each. Solve the simultaneous equations.
 
  • #5


ghostbuster25 said:
i thought i was just to put the factor into the equation. Not sure what other way to do. Is the second part correct?

Yes that is how you do it. You correctly chose to evaluate f(-1) when dividing by the (x+1) factor, but when dividing by the (x-2) factor you incorrectly chose f(-2) to evaluate. What you should be evaluating is...?
 
  • #6


Mentallic said:
Yes that is how you do it. You correctly chose to evaluate f(-1) when dividing by the (x+1) factor, but when dividing by the (x-2) factor you incorrectly chose f(-2) to evaluate. What you should be evaluating is...?

I haven't done this in a while, but shouldn't the evaluation of [tex](x+1)[/tex] be [tex]x=-16[/tex], since there is a remainder of -15 when the polynomial is divided by [tex](x+1)[/tex]
 
  • #7


Cilabitaon said:
I haven't done this in a while, but shouldn't the evaluation of [tex](x+1)[/tex] be [tex]x=-16[/tex], since there is a remainder of -15 when the polynomial is divided by [tex](x+1)[/tex]

What you're referring to is f(x)+1, rather than f(x+1)
 
  • #8


zgozvrm said:
What you're referring to is f(x)+1, rather than f(x+1)

Not quite, he is referring to f(x+1)-1

Say you had a root of 2, then (x-2) is a factor and when you evaluate f(2) you will get 0.
 

1. What are simultaneous polynomials?

Simultaneous polynomials are two or more polynomial equations that are solved at the same time. They are often used in systems of equations where there are multiple variables and equations involved.

2. How do you solve simultaneous polynomials?

The most common method for solving simultaneous polynomials is by using the elimination method or substitution method. These methods involve manipulating the equations to eliminate one variable and then solving for the remaining variable.

3. Can simultaneous polynomials have more than two variables?

Yes, simultaneous polynomials can have any number of variables as long as there are the same number of equations. However, solving for more than three variables can become more complex and may require more advanced methods.

4. What is the importance of simultaneous polynomials in science?

Simultaneous polynomials are used in many scientific fields, such as physics, chemistry, and engineering. They are used to model real-world scenarios and solve for unknown variables, making them a crucial tool in problem-solving and data analysis.

5. Are there any limitations to solving simultaneous polynomials?

While simultaneous polynomials can be used to solve a wide range of problems, there are some limitations. These equations must be linear (no exponents higher than 1) and there must be the same number of equations as variables. If these conditions are not met, different methods may need to be used to solve the equations.

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