Problems on Remainder Theorem

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).





10th Grade Math

From Problems on Remainder Theorem to HOME




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.