Remainder Theorem

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Remainder theorem : Let P(x) be any polynomial of degree greater than or equal to one and let ‘a’ be any real number. If P(x) is divided by the linear polynomial x – a, then the remainder is P(a). If P(a) = 0 then x - a is the factor of the given polynomial.

Examples :

1) Find the remainder when x⁴ - x³ - 3x² -2x +1 is divided by x – 1 .

Solution :
P(x) = x⁴ - x³ - 3x² -2x + 1 and

x-1 = 0

x = 1

So put x = 1 in P(x)

P(1) = (1) ⁴ - (1) ³ - 3(1) ² -2(1) +1

= 1 -1 – 3 -2 +1

= - 5 + 1

= - 4

So, by remainder theorem, - 4 is the remainder when x⁴ -x³ -3x² -2x +1 is divided by x – 1.

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2) Find the remainder when p(x) = 4x³ - 12x² + 14x -3 is divided by
g(x) = x -1/2.

Solution :
By remainder theorem

g(x) = x - 1/2 ⇒ x = 1/2

Put x = 1/2 in the given polynomial

p(1/2) = 4(1/2)³ - 12(1/2)² + 14(1/2) -3

= 4/8 - 12/4 + 7 - 3

= 1/2 - 3 + 4
= 1/2 + 1

p(1/2) = 3/2.

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3) Let R1 and R2 are the remainders of when the polynomials
x³ + 2x² - 5ax - 7 and x³ + ax² -12x + 6 are divided by x + 1 and x - 2 respectively. If 2R1 + R2 = 6, find the value of a.

Solution :
Let p(x) = x³ + 2x² - 5ax - 7 and q(x) = x³ + ax² -12x + 6


⇒ R1 = p(-1)
⇒ R1 = (-1)³ + 2(-1)² - 5a(-1) -7
⇒ R1 = -1 + 2 + 5a - 7
⇒ R1 = 5a - 6

And,
R2 = Remainder when q(x) is divided by x - 2
R2 = q(2)
⇒ R2 = (2)³ + a(2)² -12(2) + 6
⇒ R2 = 8 + 4a - 24 + 6
⇒ R2 = 4a - 10
As 2R1 + R2 = 6
Put the values of R1 and R2 in the above equation
2(5a - 6) + 4a - 10 = 6
10a - 12 + 4a - 10 = 6
14a - 22 = 6
14a = 28
a = 28/14
a = 2

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Polynomial

• Degree of the Polynomial
• Zeros of Polynomial
• Remainder Theorem
• Find remainder by Synthetic Division
• Rational root test in Polynomial
• Solved Examples on Polynomial

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