Finding the remainder!

After spending some mind and time I got the following result:

Given any polynomial f(x)f(x) with degree greater than or equal to 22. And you know that if it is divided by xpx-p the remainder we get is qq. And if it is divide by xqx-q then the remainder is pp, then if you divide it by (xp)(xq)(x-p)(x-q) the remainder you will get will be p+qx\boxed{p+q-x} where xx is a variable and pp and qq are constants and pq\boxed{p≠q}

Proof:

Consider any f(x)f(x), and let consider it's division by xpx-p now it is necessary that degree of f(x)f(x) is greater than or equal to 22 (see why?)

From Division algorithm of polynomials:

f(x)=(xp)g1(x)+qf(x)=(x-p)g_1(x)+q

Where g1(x)g_1(x) is any polynomial

Similarly of we divide it by xqx-q

f(x)=(xq)g2(x)+pf(x)=(x-q)g_2(x)+p

Now if we divide it by (xp)(xq)(x-p)(x-q)

f(x)=(xp)(xq)g3(x)+r(x)f(x)=(x-p)(x-q)g_3(x)+r(x)

From division algorithm of polynomial it follows that degree of r(x)r(x) is smaller than (xp)(xq)(x-p)(x-q) or r(x)=0r(x)=0

As degree of (xp)(xq)(x-p)(x-q) will be 22 therefore the highest possible degree of r(x)r(x) is11. Therefore there exists aa and bb such that r(x)=ax+br(x)=ax+b where it is not necessary that a0a≠0 as r(x)r(x) can also be a constant in that case a=0a=0 and if r(x)=0r(x)=0 then a=0a=0 and b=0b=0.

f(x)=(xp)(xq)g3(x)+(ax+b)f(x)=(x-p)(x-q)g_3(x)+(ax+b)

f(p)=(pp)(pq)g3(p)+(ap+b)=ap+bf(p)=(p-p)(p-q)g_3(p)+(ap+b)=ap+b

andand

f(q)=(qp)(qq)g3(q)+(aq+b)=aq+bf(q)=(q-p)(q-q)g_3(q)+(aq+b)=aq+b

Also as remainder of division of f(x)f(x) and (xp)(x-p) is f(p)f(p) from remainder theorem. Therefore, f(p)=qf(p)=q because both (i.e f(p)f(p) and qq) of them are the remainder of the same division, similarly f(q)=pf(q)=p therefore,

q=ap+bq=ap+b ...(1)

andand

p=aq+bp=aq+b ...(2)

if we subtract the (1) from (2) we get

pq=aqapp-q=aq-ap

pq=a(qp)p-q=a(q-p)

a=1a=-1

putting it in the equation (p=aq+bp=aq+b) we get

p=q+bp=-q+b

b=p+qb=p+q

Therefore the remainder r(x)=ax+b=x+p+q=p+qxr(x)=ax+b=-x+p+q=\boxed{p+q-x}

Note:

  • It is important to note that if p=qp=q then from the equation (pq=a(qp)p-q=a(q-p)) we can't get that a=1a=-1 because in that case pq=qp=0p-q=q-p=0 and as you can't divide anything by 00, therefore pq\boxed{p≠q} as mentioned in the statement above

Note by Zakir Husain
5 days, 2 hours ago

No vote yet
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