Remainder Theorem

Definition of Remainder Theorem:

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


In other words:

If the polynomial f(x) is divided by x - α then the remainder R is given by f(x) = (x - α) q(x) + R, where q(x) is the quotient and R is a constant (because the degree of the remainder is less than the degree of the divisor x - α).


Putting x = α, f(α) = (α - α)q(α) + R or f(α) = R

When the polynomial f(x) is divided by x - α, the remainder R = f(α) =  value of f(x) when x is α.

Solved examples on Remainder Theorem:

1. Find the remainder when x\(^{3}\) + 3x\(^{2}\) + 3x +1 is divided by

(i) x + 1

(ii) x - \(\frac{1}{2}\)

(iii) x

(iv) x + γ

(v) 5 + 2x

Solution:

(i) Let f(x) = x\(^{3}\) + 3x\(^{2}\) + 3x +1, divisor is x +1

Then by the Remainder Theorem we get,

Remainder = f(-1)

                  = (-1)\(^{3}\) + 3(-1)\(^{2}\) + 3(-1) +1

                  = -1 + 3 - 3 + 1

                  = 0

(ii) Let f(x) = x\(^{3}\) + 3x\(^{2}\) + 3x +1, divisor is x - \(\frac{1}{2}\)

Then by the Remainder Theorem we get,

Remainder = f(\(\frac{1}{2}\))

                  = (\(\frac{1}{2}\))\(^{3}\) + 3(\(\frac{1}{2}\))\(^{2}\) + 3(\(\frac{1}{2}\)) + 1

                  = \(\frac{1}{8}\) + \(\frac{3}{4}\) + \(\frac{3}{2}\) + 1

                  = \(\frac{1 + 6 + 12 + 8}{8}\)

                  = \(\frac{27}{8}\)

(iii) Let f(x) = x\(^{3}\) + 3x\(^{2}\) + 3x +1, divisor is x i.e., x - 0

Then by the Remainder Theorem we get,

Remainder = f(0)

                  = 0\(^{3}\) + 3 ∙ 0\(^{2}\) + 3 ∙  0 + 1

                  = 1

(iv) Let f(x) = x\(^{3}\) + 3x\(^{2}\) + 3x +1, divisor is x + γ

Then by the Remainder Theorem we get,

Remainder = f(-γ)

                  = (-γ)\(^{3}\) + 3(-γ)\(^{2}\) + 3(-γ) +1

                  = -γ\(^{3}\) + 3γ\(^{2}\) - 3γ +1

(v) Let f(x) = x\(^{3}\) + 3x\(^{2}\) + 3x +1, divisor is 5 + 2x

Then by the Remainder Theorem we get,

Remainder = f(-\(\frac{5}{2}\))

                   = (-\(\frac{5}{2}\))\(^{3}\) + 3(-\(\frac{5}{2}\))\(^{2}\) + 3(-\(\frac{5}{2}\)) + 1

                   = \(\frac{-125}{8}\) + \(\frac{75}{4}\) - \(\frac{15}{2}\) + 1

                   = \(\frac{-125 + 150 -60 + 8}{8}\)

                   = -\(\frac{27}{8}\)



2. If 3x\(^{2}\) - 7x + 11 is divided by x - 2 then find the remainder.

Solution:

Here p(x) = 3x\(^{2}\) - 7x + 11, divisor is x - 2

Therefore, remainder = p(2)                    [Taking x = 2 from x - 2 = 0]

                                   = 3(2)\(^{2}\) - 7(2) + 11

                                   = 12 - 14 + 11

                                   = 9

● Factorization









10th Grade Math

From Remainder Theorem to HOME




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