REMAINDER THEOREM

Polynomial of Class 9

REMAINDER THEOREM

Let p(x) be any polynomial of degree greater than or equal to 1 and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).

Proof :Let p(x) be any polynomial with degree greater than or equal to one. Suppose that when p(x) is divided by x-a, the quotient is q(x) and the remainder is r(x), i.e.,

p(x) = (x-a)q(x) + r(x)…(i)

Since the degree of x - a is one and the degree of r(x) is less than the degree of x - a , it implies that the degree of r(x) = 0This means that r(x), is a constant, say r.

∴ So, for every value of x we have r(x) = r.

Therefore equation (i) becomes p(x) = (x - a) q(x) + r

In particular, if x = a above equation gives us

p(a) = (a - a) q(a) + r

0.q(a) + r

p(a) = r

which proves the theorem.

Question Find the remainder when f(x) = x3– 6x2 + 2x – 4 is divided by g(x) = 1 – 2x.

Solution: Put g(x) = 1 – 2x = 0 ⇒ 2x = 1 ⇒ x = 1/2

REMAINDER THEOREM

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Question The polynomials ax3 + 3x2– 13 and 2x3– 5x + a are divided by x + 2 if the remainder in each case is the same, find the value of a.

Solution: p(x) = ax3 + 3x2– 13 and q(x) = 2x3– 5x + a

when p(x) & q(x) are divided by x + 2 = 0 ⇒ x = –2

p (–2) = q(–2)

⇒ a(–2)3 + 3(–2)3– 13 = 2(–2)3– 5(–2) + a

⇒ –8a + 12 – 13 = –16 + 10 + a

⇒ –9a = –5

⇒ a = 5/9 Ans.

FACTOR THEOREM:

Statement: If p(x) is a polynomial of degree n ≥ 1 and a is any real number, then (i) x - a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0 if x- a is a factor of p(x).

Proof: By the Remainder Theorem we have, p(x) = (x - a) q(x) + p(a).

(i) If p(a)= 0 then p(x) = (x - a) q(x) which indicates that x - a is a factor of p(x)

(ii) Since x - a is a factor of p(x),p(x) = (x - a) g(x) for some polynomial g(x). In this case, p(a) = ( a - a)g(a) = 0

Question Show that x + 1 an d 2x - 3 are factors of 2x3– 9x2 + x + 12.

Solution: To prove that (x + 1) and (2x - 3) are factors of 2x3– 9x2 + x + 12 it is sufficient to show that p(–1) and p(3/2)both are equal to zero.

p(–1) = 2(–1)3– 9(–1)2 + (–1) + 12 = –2 – 9 – 1 + 12 = –12 + 12 = 0

REMAINDER THEOREM

Hence, (x + 1) and (2x – 3) are the factors 2x3– 9x2 + x + 12. Ans.

REMAINDER THEOREM

From equation (i) and (ii)

–2α– 2 = –4α– 4

⇒2α = – 2 ⇒α = –1

Put α= –1 in equation (i) ⇒β = –2(–1) – 2 = 2 – 2 = 0.

Hence,α= –1 β= 0. Ans.

Division of polynomials

Long division method

Long division for polynomials works in much the same way:

Question Divide REMAINDER THEOREM by long division method

REMAINDER THEOREM

Question Divide REMAINDER THEOREM by long division method

Solution:

REMAINDER THEOREM

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