Math, asked by srikar08, 1 year ago

if f (x) is divided by (x-a)(x-b) then remainder "r" is?

Answers

Answered by santhosh6

If P(x) is divided by (x-a)(x-b), where a is not equal to b and a and b are elements of the real number set, prove that the remainder is

[(P(b)-P(a))/(b-a)] x (x-a) + P(a)

I was able to make some progress, but I am not able to fully prove this problem. Any help would be appreciated.

srikar08: We want the process

santhosh6: Suppose P(x)Px is a polynomial of degree greater than two. If P(x)Px is divided by (x−a)(x−b)xaxb, the remainder is
[Math Processing Error]

santhosh6: PaPbabxaPa
Proof:

By the remainder theorem, P(x)=q1(x)(x−a)+P(a)Pxq1xxaPa and P(x)=q2(x)(x−b)+P(b)Pxq2xxbPb. Similarily, P(x)=q3(x)(x−a)(x−b)+cx+dPxq3xxaxbcxd. We know that P(a)=ca+dPacad and P(b)=cb+dPbcbd, therefore
c=P(a)−P(b)a−b
cPaPbab
. Solving for dd in the previous equation yields
[Math Processing Error]
dPaaPaPbab
Now since the remainder is simply cx+dcxd, we obtain
[Math Processing Error]

santhosh6: cxdPaaPaPbabxPaPbabPa
which is trivially rewritten as
[Math Processing Error]
xaPaPbabPa

Now is it possible to derive such an identity for something of the form

P(x)=q1(x)(x−r1)+P(r1)Pxq1xxr1Pr1
P(x)=q1(x)(x−r1)+P(r2)Pxq1xxr1Pr2
.
.
.
P(x)=q1(x)(x−r1)+P(rn)Pxq1xxr1Prn
Therefore
P(x)=qI(x)∏i∈I(x−i)+∑|I|−1awxw+d
PxqIxiIxiI1awxwd
? Because you'd get a system of equations like
P(r1)=∑|I|−1aw(r1)w+d
Pr1I1awr1wd
P(r2)=∑|I|−1aw(rn)w+d
Pr2I1awrnwd

.
.
.
P(rn)=∑|I|−1aw(rn)w+d

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