Please solve:
f(x) = (p-1)x^3 + px^2 + r, (where p, q and r are constants), is divided by (x+2) and (x-1), the remainders are -5 and 4 respectively. If (x+1) is a factor find the values of p, q and r.
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Answer:
By the remainder theorem, f(1)=4,f(−1)=0,f(−2)=5
This gives the simultaneous equations:
(p−1)+p+q+r=4
1−p+p−q+r=0
8–8p+4p−2q+r=5
Simplifying :
2p+q+r=5
q−r=1
4p+2q−r=3
Replacing r=q−1
from the second equation in the others:
p+q=3
4p+q=2
Subtracting the first equation from the second:
3p=−1,p=−13
q=3−−13=103
r=103−1=73
f(x)=−43x3+−13x2+103x+73
And the factors include x+1
by the remainder theorem. For the other factors, we set f(x)=0
and factor:
4x3+x2−10x−7=0
(x+1)(4x2–3x−7)=0
(x+1)2(4x−7)=0
f(x)=−13(x+1)2(4x−7)
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