A Ambility Active Member Joined Dec 22, 2014 Messages 336 Gender Male HSC 2016 Feb 28, 2016 #1 How would you do this?
leehuan Well-Known Member Joined May 31, 2014 Messages 5,805 Gender Male HSC 2015 Feb 28, 2016 #2 P(x)=(x-1)(x+2)(ax+b) If you sub for P(-1)=4 and P(2)=28 you're gonna get a simultaneous equation in a and b which allows you finish the question off.
P(x)=(x-1)(x+2)(ax+b) If you sub for P(-1)=4 and P(2)=28 you're gonna get a simultaneous equation in a and b which allows you finish the question off.
A Ambility Active Member Joined Dec 22, 2014 Messages 336 Gender Male HSC 2016 Feb 28, 2016 #3 Shouldn't it be something like this though? P(x) = c(x-1)(x+2)(ax+b) Because we don't know the coefficient of the whole polynomial?
Shouldn't it be something like this though? P(x) = c(x-1)(x+2)(ax+b) Because we don't know the coefficient of the whole polynomial?
leehuan Well-Known Member Joined May 31, 2014 Messages 5,805 Gender Male HSC 2015 Feb 28, 2016 #4 Ambility said: Shouldn't it be something like this though? P(x) = c(x-1)(x+2)(ax+b) Because we don't know the coefficient of the whole polynomial? Click to expand... Note that the a catered for it. P(x)=(x-1)(x+2)(cax+bc) P(x)=(x-1)(x+2)(dx+e) You're thinking about: P(x)=a(x-1)(x+2)(x+b)
Ambility said: Shouldn't it be something like this though? P(x) = c(x-1)(x+2)(ax+b) Because we don't know the coefficient of the whole polynomial? Click to expand... Note that the a catered for it. P(x)=(x-1)(x+2)(cax+bc) P(x)=(x-1)(x+2)(dx+e) You're thinking about: P(x)=a(x-1)(x+2)(x+b)
A Ambility Active Member Joined Dec 22, 2014 Messages 336 Gender Male HSC 2016 Feb 28, 2016 #5 leehuan said: Note that the a catered for it. P(x)=(x-1)(x+2)(cax+bc) P(x)=(x-1)(x+2)(dx+e) You're thinking about: P(x)=a(x-1)(x+2)(x+b) Click to expand... Thanks
leehuan said: Note that the a catered for it. P(x)=(x-1)(x+2)(cax+bc) P(x)=(x-1)(x+2)(dx+e) You're thinking about: P(x)=a(x-1)(x+2)(x+b) Click to expand... Thanks