What Are the Correct Values of a and b in the Polynomial Problem?

[ghostbuster25]

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

 

[Mentallic]

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)

 

[ghostbuster25]

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

 

[zgozvrm]

First, divide [itex](x-2)[/tex] into [itex]4x^3+ax^2+bx+6[/tex]<br /> The remainder will have a- and b-terms in it.<br /> 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.<br /> <br /> 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.<br /> <br /> Now, divide [itex](x+1)[/tex] into [itex]4x^3+ax^2+bx+6[/tex].<br /> Again, you'll get a remainder with a- and b-terms in it.<br /> Set <i>this</i> remainder to -15.<br /> <br /> You now have 2 equations with 2 variables (a and b) each. Solve the simultaneous equations.[/itex][/itex][/itex][/itex][/itex][/itex]

 

[Mentallic]

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...?

 

[Cilabitaon]

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 $$(x+1)$$ be $$x=-16$$, since there is a remainder of -15 when the polynomial is divided by $$(x+1)$$

 

[zgozvrm]

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

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

 

[Mentallic]

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.