# 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).

● Factorization

10th Grade Math

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