ICSE Class 10 Maths Chapter 08 Remainder and Factor Theorems

Remainder And Factor Theorems

A Basic Concept

In equation \(f(x) = 2x^2 - 5x - 7\), \(f(x)\) is said to be a function of variable \(x\) as the value of \(f(x)\) depends on the value of \(x\). The following examples prove this statement.

\(f(x) = 2x^2 - 5x - 7 \Rightarrow\)

(i) if \(x = 3\), \(f(3) = 2 \times (3)^2 - 5 \times 3 - 7 = 18 - 22 = -4\)

(ii) if \(x = -3\), \(f(-3) = 2(-3)^2 - 5(-3) - 7 = 18 + 15 - 7 = 26\) and so on.

In the same way: in \(f(y) = 2y^3 - 3y + 1\), \(f(y)\) is a function of variable \(y\) as the value of \(f(y)\) depends on the value of \(y\), e.g.,

\(f(y) = 2y^3 - 3y + 1 \Rightarrow\)

(i) if \(y = 5\), \(f(5) = 2 \times 5^3 - 3 \times 5 + 1 = 250 - 15 + 1 = 236\).

(ii) if \(y = -2\), \(f(-2) = 2 \times (-2)^3 - 3 \times -2 + 1 = -16 + 6 + 1 = -9\) and so on.

Find the remainder obtained on dividing \(f(x) = x^2 - 5x + 8\) by \(x - 2\).

Now, find \(f(2)\)

\(f(x) = x^2 - 5x + 8\)

\(\Rightarrow f(2) = (2)^2 - 5 \times 2 + 8\)

\(= 4 - 10 + 8 = 2\)

\(= \) The remainder when \(x^2 - 5x + 8\) is divided by \(x - 2\).

Again, find the remainder obtained on dividing \(f(x) = 6x^3 - 3x^2 + 8x - 5\) by \(x + 3\)

Now, find \(f(-3)\)

\(f(x) = 6x^3 - 3x^2 + 8x - 5\)

\(\Rightarrow f(-3) = 6(-3)^3 - 3(-3)^2 + 8(-3) - 5\)

\(= -162 - 27 - 24 - 5\)

\(= -218\)

\(= \) The remainder when \(6x^3 - 3x^2 + 8x - 5\) is divided by \(x + 3\)

It is clear from the examples, given above, that:

1. when \(f(x)\) is divided by \(x - 2\), the remainder = the value of \(f(2)\).

2. when \(f(x)\) is divided by \(x + 3\), the remainder = the value of \(f(-3)\)

The method of finding the remainder without actually performing the process of division is called Remainder Theorem.

Remainder Theorem

If \(f(x)\), a polynomial in \(x\), is divided by \((x - a)\), the remainder = \(f(a)\)

e.g. If \(f(x)\) is divided by \((x - 3)\), the remainder is \(f(3)\).

For finding the remainder, using Remainder Theorem:

Step 1: Put the divisor equal to zero and solve the equation obtained to get the value of its variable.

Step 2: Substitute the value of the variable, obtained in step 1, in the given polynomial and simplify it to get the required remainder.

Find the remainder when \(x^2 - 8x + 4\) is divided by \(2x + 1\).

Solution:

Step 1: \(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)

Step 2: Required remainder = Value of given polynomial \(x^2 - 8x + 4\) at \(x = -\frac{1}{2}\)

\(\therefore \) Remainder = \(\left(-\frac{1}{2}\right)^2 - 8\left(-\frac{1}{2}\right) + 4\)

\(= \frac{1}{4} + 4 + 4 = 8\frac{1}{4}\) Ans.

Teacher's Note

Understanding remainders is like checking your change at a store - if you divide a total amount by the number of equal payments, any leftover amount (remainder) tells you what's not evenly distributed.

Find the value of 'a' if the division of \(ax^3 + 9x^2 + 4x - 10\) by \(x + 3\) leaves a remainder 5.

Solution:

\(x + 3 = 0 \Rightarrow x = -3\)

Given, remainder is 5; therefore:

The value of \(ax^3 + 9x^2 + 4x - 10\) at \(x = -3\) is 5

\(\Rightarrow a(-3)^3 + 9(-3)^2 + 4(-3) - 10 = 5\)

\(\Rightarrow -27a + 81 - 12 - 10 = 5\) or \(a = 2\) Ans.

When the polynomial \(2x^3 - kx^2 + (5k - 3)x - 8\) is divided by \(x - 2\), the remainder is 14. Find the value of 'k'.

Solution:

\(x - 2 = 0 \Rightarrow x = 2\)

Given, remainder is 14, therefore:

\(\Rightarrow 2(2)^3 - k(2)^2 + (5k - 3) \times 2 - 8 = 14\)

\(\Rightarrow 16 - 4k + 10k - 6 - 8 = 14\)

\(\Rightarrow 6k = 12\) and \(k = 2\) Ans.

Teacher's Note

This is like reverse-engineering a recipe - if you know the final result (remainder), you can work backwards to find the missing ingredient (the constant k).

The polynomials \(3x^3 - ax^2 + 5x - 13\) and \((a + 1)x^2 - 7x + 5\) leave the same remainder when divided by \(x - 3\). Find the value of 'a'.

Solution:

\(x - 3 = 0 \Rightarrow x = 3\)

Since, the given polynomials leave the same remainder when divided by \(x - 3\)

Value of polynomial \(3x^3 - ax^2 + 5x - 13\) at \(x = 3\) is the same as the value of polynomial \((a + 1)x^2 - 7x + 5\) at \(x = 3\)

\(\Rightarrow 3(3)^3 - a(3)^2 + 5 \times 3 - 13 = (a + 1)(3)^2 - 7 \times 3 + 5\)

\(\Rightarrow 81 - 9a + 15 - 13 = 9a + 9 - 21 + 5\)

\(\Rightarrow 18a = 90\) and \(a = 5\) Ans.

When \(f(x) = x^3 + ax^2 - bx - 8\) is divided by \(x - 2\), the remainder is zero and when divided by \(x + 1\), the remainder is -30. Find the values of 'a' and 'b'.

Solution:

Since, \(x - 2 = 0 \Rightarrow x = 2\)

And, given that on dividing \(f(x) = x^3 + ax^2 - bx - 8\) by \(x - 2\), the remainder = 0

\(\Rightarrow f(2) = 0\)

\(\Rightarrow (2)^3 + a(2)^2 - b(2) - 8 = 0\) i.e. \(8 + 4a - 2b - 8 = 0\)

\(\Rightarrow 4a - 2b = 0\) i.e. \(2a - b = 0\) ....I

Again, given that on dividing \(f(x) = x^3 + ax^2 - bx - 8\) by \(x + 1\), the remainder = -30

\(\Rightarrow f(-1) = -30\) \([x + 1 = 0 \Rightarrow x = -1]\)

\(\Rightarrow (-1)^3 + a(-1)^2 - b(-1) - 8 = -30\) i.e. \(-1 + a + b - 8 = -30\)

\(\Rightarrow a + b = -21\) ....II

On solving equations I and II, we get: \(a = -7\) and \(b = -14\) Ans.

What number should be added to \(2x^3 - 3x^2 + x\) so that when the resulting polynomial is divided by \(x - 2\), the remainder is 3?

Solution:

Let the number added be \(k\) so the resulting polynomial is \(2x^3 - 3x^2 + x + k\)

Given, when this polynomial is divided by \(x - 2\), the remainder = 3

\(\Rightarrow 2(2)^3 - 3(2)^2 + 2 + k = 3\) \([x - 2 = 0 \Rightarrow x = 2]\)

\(\Rightarrow 16 - 12 + 2 + k = 3\) i.e., \(k = -3\)

\(\therefore \) The required number to be added = -3 Ans.

Teacher's Note

Adding a number to balance an equation is like adjusting the weight on a scale to make it level - you need to find exactly how much to add or remove to get the desired result.

Factor Theorem

When a polynomial \(f(x)\) is divided by \(x - a\), the remainder = \(f(a)\). And, if remainder \(f(a) = 0\); \(x - a\) is a factor of the polynomial \(f(x)\).

For example:

Let \(f(x) = x^2 - 5x + 6\) be divided by \(x - 3\);

then remainder = \(f(3)\)

\(= (3)^2 - 5 \times 3 + 6 = 0\)

\(\therefore\) Remainder = 0

\(\Rightarrow x - 3\) is a factor of \(f(x) = x^2 - 5x + 6\)

Determine whether \(x - 1\) is a factor of \(x^6 - x^5 + x^4 + x^3 - x^2 - x + 1\) or not?

Solution:

\(x - 1 = 0 \Rightarrow x = 1\)

\(\therefore\) When given polynomial is divided by \(x - 1\), the remainder

\(= (1)^6 - (1)^5 + (1)^4 + (1)^3 - (1)^2 - (1) + 1\)

\(= 1 - 1 + 1 + 1 - 1 - 1 + 1\)

\(= 4 - 3 = 1\), which is not equal to zero.

\(\therefore x - 1\) is not a factor of the given polynomial. Ans.

If \(x - 2\) is a factor of \(x^2 - 7x + 2a\), find the value of a.

Solution:

\(x - 2 = 0 \Rightarrow x = 2\)

Since, \(x - 2\) is a factor of polynomial \(x^2 - 7x + 2a\)

\(\Rightarrow\) Remainder = 0 \(\Rightarrow (2)^2 - 7(2) + 2a = 0 \Rightarrow a = 5\) Ans.

Find the value of 'k' if \((x - 2)\) is a factor of \(x^3 + 2x^2 - kx + 10\).

Hence, determine whether \((x + 5)\) is also a factor. [2011]

Solution:

\(x - 2\) is a factor and \(x - 2 = 0 \Rightarrow x = 2\)

\(\therefore\) The value of given expression \(x^3 + 2x^2 - kx + 10\) is zero at \(x = 2\)

i.e. remainder = 0

\(\Rightarrow (2)^3 + 2(2)^2 - k \times 2 + 10 = 0\)

\(\Rightarrow 8 + 8 - 2k + 10 = 0\)

\(\Rightarrow k = 13\) Ans.

On substituting \(k = 13\), the given expression becomes \(x^3 + 2x^2 - 13x + 10\).

Now to check whether \((x + 5)\) is also a factor or not,

find the value of the given expression for \(x = -5\). \([\therefore x + 5 = 0 \Rightarrow x = -5]\)

\(\therefore x^3 + 2x^2 - 13x + 10\) (at \(x = -5\))

\(= (-5)^3 + 2(-5)^2 -13(-5) + 10\)

\(= -125 + 50 + 65 + 10 = -125 + 125 = 0\)

Since, the remainder is 0 \(\Rightarrow (x + 5)\) is a factor Ans.

Given that \(x + 2\) and \(x - 3\) are factors of \(x^3 + ax + b\); calculate the values of a and b.

Solution:

Given, \(x + 2\) is a factor of \(x^3 + ax + b\);

\(\Rightarrow (-2)^3 + a(-2) + b = 0\) \([x + 2 = 0 \Rightarrow x = -2]\)

\(\Rightarrow -2a + b = 8\) .....I

Again, given that:

\(x - 3\) is a factor of \(x^3 + ax + b\);

\(\Rightarrow (3)^3 + a(3) + b = 0\) \([x - 3 = 0 \Rightarrow x = 3]\)

\(\Rightarrow 3a + b = -27\) .....II

On solving equations I and II, we get \(a = -7\) and \(b = -6\) Ans.

Polynomial \(x^3 - ax^2 + bx - 6\) leaves remainder -8 when divided by \(x - 1\) and \(x - 2\) is a factor of it. Find the values of 'a' and 'b'.

Solution:

On dividing by \(x - 1\), the polynomial \(x^3 - bx^2 + bx - 6\) leaves remainder -8

\(\Rightarrow (1)^3 - a(1)^2 + b(1) - 6 = -8\) \([x - 1 = 0 \Rightarrow x = 1]\)

\(\Rightarrow -a + b = -3\)

i.e. \(a - b = 3\) .....I

(\(x - 2\)) is a factor of polynomial \(x^3 - bx^2 + bx - 6\)

\(\Rightarrow (2)^3 - a(2)^2 + b(2) - 6 = 0\) \([x - 2 = 0 \Rightarrow x = 2]\)

\(\Rightarrow 8 - 4a + 2b - 6 = 0\)

i.e. \(2a - b = 1\) .....II

On solving equations I and II, we get:

\(a = -2\) and \(b = -5\) Ans.

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