We will discuss here how to solve the problems on Remainder Theorem.
1. Find the remainder (without division) when 8x\(^{2}\) +5x + 1 is divisible by x - 10
Solution:
Here, f(x) = 8x\(^{2}\) + 5x + 1.
By remainder Theorem,
The remainder when f(x) is divided by x – 10 is f(10).
2. Find the remainder when x\(^{3}\) - ax\(^{2}\) + 6x - a is divisible by x - a.
Solution:
Here, f(x) = x\(^{3}\) - ax\(^{2}\) + 6x - a, divisor is (x - a)
Therefore, remainder = f(a) , [ Taking x = a from x - a = 0]
= a\(^{3}\) - a ∙ a\(^{2}\) + 6 ∙ a - a
= a\(^{3}\) -a\(^{3}\) + 6a - a
= 5a.
3. Find the remainder (without division) when x\(^{2}\) +7x - 11
is divisible by 3x - 2
Solution:
Here, f(x) = x\(^{2}\) + 7x – 11 and 3x - 2 = 0 ⟹ x = \(\frac{2}{3}\)
By remainder Theorem,
The remainder when f(x) is divided by 3x - 2 is f(\(\frac{2}{3}\)).
Therefore, remainder = f(\(\frac{2}{3}\)) = (\(\frac{2}{3}\))\(^{2}\) + 7 ∙ (\(\frac{2}{3}\)) - 11
= \(\frac{4}{9}\) + \(\frac{14}{3}\) - 11
= -\(\frac{53}{9}\)
4. Check whether 7 + 3x is a factor of 3x\(^{3}\) + 7x.
Solution:
Here f(x) = 3x\(^{3}\) + 7x and divisor is 7 + 3x
Therefore, remainder = f(-\(\frac{7}{3}\)), [Taking x = -\(\frac{7}{3}\) from 7 + 3x = 0]
= 3 ∙ (-\(\frac{7}{3}\))\(^{3}\) + 7(-\(\frac{7}{3}\))
= -3 × \(\frac{343}{27}\) - \(\frac{49}{3}\)
= \(\frac{-343 - 147}{9}\)
= \(\frac{-490}{9}\)
≠ 0
Hence, 7 + 3x is not a factor of f(x) = 3x\(^{3}\) + 7x.
5. Find the remainder (without division) when 4x\(^{3}\) - 3x\(^{2}\) + 2x - 4 is divisible by x + 2
Solution:
Here, f(x) = 4x\(^{3}\) - 3x\(^{2}\) + 2x - 4 and x + 2 = 0 ⟹ x = -2
By remainder Theorem,
The remainder when f(x) is divided by x + 2 is f(-2).
Therefore, remainder = f(-2) = 4(-2)\(^{3}\) - 3 ∙ (-2)\(^{2}\) + 2 ∙ (-2) - 4
= - 32 - 12 - 4 - 4
= -52
6. Check whether the polynomial: f(x) = 4x\(^{3}\) + 4x\(^{2}\) - x - 1 is a multiple of 2x + 1.
Solution:
f(x) = 4x\(^{3}\) + 4x\(^{2}\) - x - 1 and divisor is 2x + 1
Therefore, remainder = f(-\(\frac{1}{2}\)), [Taking x = \(\frac{-1}{2}\) from 2x + 1 = 0]
= 4 ∙ (-\(\frac{1}{2}\))\(^{3}\) + 4(-\(\frac{1}{2}\))\(^{2}\) - (-\(\frac{1}{2}\)) -1
= -\(\frac{1}{2}\) + 1 + \(\frac{1}{2}\) - 1
= 0
Since the remainder is zero ⟹ (2x + 1) is a factor of f(x). That is f(x) is a multiple of (2x + 1).
● Factorization
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