Page 23 - 11-Math-4 Quadratic Equations
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Example 1: Find the remainder when the polynomial x3 + 4x2 - 2x + 5 is divided by x - 1.
Solution: Let f(x) = x3 + 4x2 - 2x + 5 and x - a = x - 1 ⇒ a = 1
Remainder = f(1)
(By remainder theorem)
= (1)3 +4(1)2 - 2(1) + 5
= 1+4-2+5
= 8
Example 2: Find the numerical value of k if the polynomial x3 + kx2 - 7x + 6 has a remainder
of - 4, when divided by x + 2.
Solution: Let f(x) = x3 + kx2 - 7x + 6 and x - a = x + 2, we have, a = -2
Remainder = f(-2) (By remainder theorem)
= (-2)3 + k(-2)2 - 7(-2) + 6
= -8 + 4k + 14 + 6
= 4k + 12
Given that remainder = - 4
∴ 4k + 12 = - 4
⇒ 4k = -16
⇒ k=-4
Factor Theorem: The polynomial x - a is a factor of the polynomial f ( x ) if and only if
f ( a ) = 0 i.e.; (x - a) is a factor of f ( x ) if and only if x = a is a root of the polynomial equation
f ( x ) = 0.
Proof: Suppose g(x) is the quotient and R is the remainder when a polynomial f (x) is divided
by x - a, then by Remainder Theorem
f (x) = ( x- a) g(x) + R
Since f (a) = 0 ⇒ R = 0
∴ f (x) = ( x- a) g(x)
∴ ( x- a) is a factor of f(x).
Conversely, if ( x- a) is a factor of f(x), then
R = f (a) = 0
which proves the theorem.
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