## Function and Polynomials

Subject: Optional Mathematics

#### Overview

If f: A → B be a function from set A to set B and g: B → C be the function set B to set C, then the new function from set A to set C is called composite function of F and G. It is denoted by gof. The algebraic expressions in which the power of the variables is whole numbers are called polynomials. Rules of polynomials: 1 . Multiply each term of one polynomial by each term of the order. 2 . Arrange the terms in ascending or descending order.

### FUNCTION

Let A and B be two non-empty sets, then, every subset of cartesian product A$\times$B is a relation from A to B. Thus, a relation in which every element of set A is related (or associated) with a unique element of set B is said to be a function from A to B. Such as function is denoted by f : A→ B, which is read as "f is a function from A to B".

Symbolically, it follows that f: A → B if;

(i) f ⊆A×B

(ii) for each given x ∈ A, there exist a unique y∈B such that (x, y) ∈ f.

If an element y of B is associated with an element x of A, then y is an image of x under f and x is called the pre-image of y under f. We can also write y=f(x) and read f(x) as "f of x" or "f at x". A function is usually denoted by f, g, F, G, etc. A function is a relation in which, no two different ordered pairs have the same first component.

#### Composite Function

Let f and g be the function from A to B and from B to C, respectively, where A= {a1,b2,c3}, B = {c3,d4,e5} and C = {e5, f6, g7,h8,i9} and these are defined as follows:

Here, if f : A →B, a1 maps into c3 ; b2 maps into d4 ; and c3 maps into e5.

If g: B→ C, c3 maps into f6 ; d4 maps into g7 ; and e5 maps into h8.

thus we may write c3= f (a1), d4 = f(b2), e5= f(c3) and f6= g(c3), g7 = g(d4), h8 = f(e5)

Then, we have f6 = g(c3) = g [f(a1)],

g7 = g(d4) = g[f(b2)]

and h8=g(e5) = g[f(e3)]

this defines a set of ordered pairs of a function from A into C. This function is called the (product) composite function of g and f. it is denoted by gof or simply by gof. the nation gof indicates that f is applied first and then g. Thus, the arrow diagram of the new function is alongside.

#### Inverse Function

Let f: A→ B be a function from A to B defined by the following arrow.

thus, the function f = { (a,1), (b,2) (c,3) } is one -one into where the domain of f = {a, b, c} and the range of f= { 1,2,3 }.

If we interchange the domain set of 'f' into range set and range set of into domain set, the set of ordered pairs will become {(1, a), (2, b), (3, c)}. Let this function e g. Then the function.

g= {(1, a), (2, b) (3, c) } is again a one-one onto, where the domain of g= {1,2,3} and the range of g={a,b,c}

the arrow diagram of this function g is as follows.

in such case, the function g is called the inverse function of and vice-versa. the inverse of the function f(x) is written as f-1read as 'f inverse ' and we have y=f(x) if and only if x = $f^{-1}$ (y) for every x∈D(f) and every y∈R(f). let u consider a many-one function defined by the following arrow diagram.

Thus, the function h= {(1, 2), (2,2), (3, 3) }

Let, we interchange the domain set and range set of h. we will get a relation R which is given below.

R= {(2, 1), (2, 2), (3, 3) }. Then the arrow diagram of this relation is given in the adjoining figure.

In relation R, two ordered pairs (2,1) and (2, 2) have the same first element 2. So, R is not a function.

It is concluded that a function f: A→ B will have its inverse function $f^{-1}$ B→ A if and only if f is a one -one onto function.

Definition: If f: A→ B is a one to one onto function from A to B, then there exist a function g: B→ A such that the range of f is the domain of g is called the inverse of 'f' denoted by $f{-1}$ (f inverse) such that y= f(x) if and only x=$f{-1}$ (y) for every x∈D (f) and every y∈R(f).

#### Simple Algebraic Functions

A function that can be defined as the root of a polynomial equation is known as an algebraic function. we shall discuss some example of functions defined on the set R real number onto itself and these functions f(x) are defined by means of an equation. for instance the algebraic functions. \begin{align}f: R→R\end{align}

$$f(x)= a_ox^n +a_1x^n-1+……………+ a_n-1 x+a_n$$

For all x€R where the right side is a polynomial of degree n.

We shall discuss some particular causes of this equation.

Constant Function: For n=0, we have f(x)=ao. it is usually denoted by f(x)=c.

In other words, a function is said to be a constant function if all its function values are the same. The graph of constant function is a straight line parallel to the x-axis at a given distance

Linear functions: For n=1, we have $$f(x) = a_ox + a_1$$. It is usually denoted by y = ax+b

In another word, a function is said to be a linear function if the polynomial is degree one. The graph of it is a straight line with slope m=a and y-intercept = for instance, y=x+2 is an equation of a straight line.

Identity function: if a=1 and b=0 in the linear function f(x) = ax +b, then we have

$$F(x) = x$$ for all $$x€R$$

This function is called identity function and its graph is shown below. It bisects the angle between the axes of coordinates.

Quadratic function: for n = 2, we have $$f(x)= a_1x^2 + a_1x + a_2.$$ It is written as $$f(x)=ax^2 + bx + c$$

This is a quadratic function which is polynomial of degree 2 and its graph is a parabola.

For instance, $$f(x) = x^{2} + x -2, f(x) = 4x^2, f(x) = x^2 + 2$$, etc are quadratic functions.

The graph of these functions is shown below.

Cubic function : for n= 3, we have$(x) f(x) = a_ox^3 +a_1x^2 + a_2x + a_3$ It is written as $f(x)=ax^3 + bx^2 +cx + d$

This is a cubic function which is polynomial of degree 3. For instance $$f(x) = x^3$$ is a cubic function whose graph is shown alongside

#### Trigonometric functions

A function defined as the function which associates each angle with the definite real number.So, the domain of the trigonometric function is the set of angles and its co-domain is the set of real numbers. traditionally trigonometric functions are defined for angles of a triangle.But these trigonometric functions can be defined for angles of any magnitude.

The trigonometric ratios for angles such as 300, 450, 600, etc. can be calculated with the help of elementary plane geometry. The following table shows the values of trigonometrical ratios from 00 to 3600.

 θ 00 300 450 600 900 1200 1350 1500 1800 Sinθ 0 $\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ 1 $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2}$ 0 Cosθ 1 $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt 2}$ $\frac{1}{2}$ 0 $-\frac{1}{2}$ $-\frac{1}{\sqrt{2}}$ $-\frac{\sqrt{3}}{2}$ -1 Tanθ 0 $\frac{1}{\sqrt{3}}$ 1 $\sqrt 3$ ∞ $-\sqrt 3$ -1 $-\frac{1}{\sqrt{3}}$ 0

### POLYNOMIALS

Polynomial is an algebraic expression consist of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. 2x, 5x2 + 3, 2a3 - b2 + 3x - 2, etc are the examples of polynomials.

Simple Operations on Polynomials

#### Multiplication of polynomials

In order to multiply two polynomials, following steps are used.

(i) Multiply each term of one polynomial by each term of the other and

(ii) Then add (or subtract) the like terms thus obtained.

(iii) Simplify and arrange the like terms in ascending order or descending order.

In multiplying two polynomials, the law of indices $x^m×x^n=x^{m+n}$ is applied.

it is noted that the degree of the product of two polynomials is equal to the sum of the degrees of the polynomial factors.

Worked out with examples

Example 1.

If $f(x)=3x^3v^4\: and\:g(x)=5x^2y^3$, find $f(x).g(x)$

Solution:

\begin{align*}f(x)\times g(x) &=3x^3y^4×5^2y^3=(3×5)\:(x^4×y^3) \\ &=15.x^{3+2}.y^{4+3}=15x^4y^7\end{align*}

Example 2.

If $f(x)=x^2-3x+2$ and $g(x)=x^3-x^2+2x+4$, find $f(x).g(x)$.

Solution:

##### Horizantal method:

\begin{align*} f(x)\times g(x)&=(x^2-3x+2)(x^3-x^2+2x+4) \\ &=x^2(x^3-x^2+2x+4)-3x(x^3-x^2+2x+4)+2(x^3-x^2+2x+4) \\ &=x^5-x^4+2x^3+4x^2-3x^4+3x^3-6x^2-6x^2-12x+2x^3-2x^2+4x+8 \\ &=x^5+(-1-3)x^4+(2+3+1)x^3+(4-6-2)x^2 (-12+4)x+8 \\ &=x^5-4x^4+7x^3-4x^2-8x+8\end{align*}

##### Vertical Method:

\begin{align*}\frac{\frac{x^3-x^2+2x+4\\\times\:x^2-3x+2}{2x^3-2x^2+4x+8\\-3x^4+3x^3-6x^2-12x\\x^5-x^4+2x^3-4x^2\\}}{x^5-4x^4+7x^3-4x^2-8x+8}\\ \end{align*}

#### Division of Polynomials

We know from arithmetic how to divide ad integer by another smaller integer. If 30 is divided by 7, the quotient is 4 and the remainder is 2. i.e., \begin{align*}7)30(4 \\ \frac{-28}{2}\end{align*}

Here, we observe that $30=7×4+2.$

this relation ca be stated as follows.

Dividend= Divisor × Quotient + Reminder

Similarly, we can divide polynomials.

let $f(x)$ and g(x) be two polynomials such that g(x) is a polynomial of smaller degree than that of $f(x)$ and g(x)≠0. Then, there exist unique polynomials Q(x) and R(x) such that

$f(x)=g(x).Q(x)+R(x)$

where $F(x)$=dividend, g(x)=divisor, Q(x) is quotient and R(x) is remainder.

If R(x) = 0, then the divisor g(x) is a factor of the dividend f(x). The other factor of f(x) is the quotient Q(x).

The relation f(x) = g(x). Q(x) + R(x).

The following steps are used to divide a polynomial by the other:

(i) Arrange the divided f(x) and divisor g(x) in standard form i.e. generally descending powers of variable x.

(ii) Divided the first term divided f(x) by the first term of divisor g(x) to get the first term of quotient Q(x).

(iii) Multiply each term of divisor g(x) by the first term of quotient Q(x) obtained in step (ii) and subtract the product so obtained from the dividend f(x).

(iv) Take the remainder obtained in step (iii) as new dividend and continue the above process until the degree of the remainder is less than of the divisor.

#### Remainder Theorem

Statement: If a polynomial f(x) is divided by x - a, then the remainder is f(a).

Proof: If we divided f(x) by x - a, then we get Q(x) as quotient and R as remainder.

Then, f(x) = (x - a). Q(x) + R

Put x = a. Then,

or, f(a) = (a - a) Q (a) + R

or, R = f(a)

Hence, remainder = f(a) = the value of polynomial f(x) = a

#### Factor theorem

Statement: if f (x) is a polynomial and a is real number, then (x - a) is a factor of f(x) if f(a) = 0

Proof:If we divide f(x) by x-a, then we get Q(x) as quotient and R as remainder.

Then, f(x) = (x - a). Q(x) + R ...........(i)

Put x = a. Then

f(a) = (a - a). Q(a) + R

or, f(a) = R

When f(a) = 0. Then R = 0.

Putting the value of R in (i) we get,

f(x) = (x - a). Q(x)

So, (x - a) is a factor of f(x).

Hence, (x - a) is a factor of f(x) if f(a) = 0.

#### Synthetic Division

Synthetic division is the process which helps us to find the quotient and remainder when a polynomial f(x) is divided by x - a.

Application of synthetic division

Let Q(x) and R be the quotient and remainder when a polynomial f(x) is divided by ax -b.

Then, f(x) = (ax - b). Q(x) + R = a(x - $\frac{b}{a}$). Q(x) + R .

Where a.Q(x) = g(x)

or, Q(x) = $\frac{1}{a}$ g(x).

Here, g(x) and R are the quotient and reminder when f(x) is divide by (x - $\frac{b}{a}$).

This result leads us to conclude that process of synthetic division discussed earlier is also useful to find out the quotient and remainder when f(x) is divided by (ax - b)

#### Factorization of a polynomial

Factor theorem and the synthetic division are very useful to find the factors of a polynomial.

Let us see the following example:

Factorize: x2 + x - 2

Constant term of this polynomial is 4 and the possible factors of 4 are: ±1, ±2.

Since the degree of f(x) is 2, so there will be at most two factors.

When x = 1, f(1) = 1+1-2 =0

(x-1) is a factor.

When x = -1, f(1) = -1-1-2 =-4

(x+1) is not a factor.

when x = 2, f(2) = 4+2-2 = 4

(x+2) is not a factor

When x = -2, f(1) = 4-2-2 = 0

(x+2) is a factor.

∴ (x-1) and (x+2) is a factor.

∴ x2 + x - 2 = (x-1)(x+2).

But instead of finding all the factors by using factor theorem, the synthetic division can be used after getting one with the help of factor theorem.

#### Polynomial Equation

Let f(x) = anxn + an-1xn-1 + ..... + ao be the polynomial in x. Then f(x) = 0 is called a polynomial equation in x.

ax + b = 0 is a linear equation.

zx2+ bx + c = 0 is a quadratic equation.

ax$^3$ + bx2 + cx + d = 0 is a cubic equation

ax$^4$ + bx$^3$ + cx2 + dx + e = 0 is a big quadratic equation.

( Here, a,b,c,d,e are the real numbers).

If α is a real number such that f(α) = 0, thenα is called a root of the polynomial equation f(x) = 0.

##### Things to remember

Functions:

Let A and B be two non-empty sets, then, every subset of cartesian product A$\times$B is a relation from A to B. Thus, a relation in which every element of set A is related ( or associated ) with a unique element of set B is said to be a function from A to B. Such as function is denoted by f : A→ B, which is read as "f is a function from A to B".

Polynomials:

Polynomial is an algebraic expression consist of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. An example of a polynomial of a single indeterminate x is x2 − 4x + 7.

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### The Factor Theorem and The Remainder Theorem

f= {(1,3), (0,0), (-1,-3)}, g= {(0,2), (-3,-1), (3,5)}, the above relation shown in the mapping diagram as follow:

gof= g(3) = 5

gof = g(0) =2

gof= g(-3) = -1

gof = g {(0,2), (-1,-1), (1,5)} Ans

f = {(-2,4), (-1,8), (0,0), (1,4), (2,6)},

g = {(o,-2), (4,0), (6,1), (8,-2)}

The mapping diagram is shown below:

gof(-2) = g(4) = 0

gof(-1) = g(8) = -2

gof(0) = g(0) = 0

gof(1) = g(4)=0

gof(2) = g(6) = 1

gof= {(-2,0), (-1,-2), (0,-2) , (1,0), (2,1)} Ans

f = {(p,1), (q,2), (r,3), (s,4)}, g = {(1,5), (2,8), (3,-3), (4,6)}

The mapping diagram is shown below:

fog = {(5,p), (8,q), (-3,r), (6,s)} Ans

f = {(x,3), (y,4), (z,7)}, g = {(3,a), (4,b), (7,c)}

The mapping diagram is shown below:

fog = {(a,x), (b,y), (c,z)} Ans

f(x) = x3 and g(x) = x2

fg(x)

= f(x2) [$\because$ g(x) = x2]

= (x2)3 [$\because$ f(x) = x3]

= x6Ans

f(x) =$\frac 1x$

ff(x) = f $\frac 1x$ [$\because$ f(x) = $\frac 1x$]

= $\cfrac{1}{ \cfrac{1}{x}}$ =1× $\frac x1$ = x

∴ ff(2) = 2 Ans

Here, f(x) = 3x2 + x - 28 and g(x) = 6-x

f(x)⋅ g(x)

= (3x2 + x - 28) (6-x)

= 18x2 + 6x - 168 -3x3-x2 + 28x

= -3x3 + 17x2+ 34x - 168 Ans

Here,

f = {(2,$\frac 12$), (3, $\frac13$), (4,$\frac 14$)}

Range = { $\frac 12$, $\frac 13$, $\frac 14$}

Inverse function (f-1) = {($\frac 12$, 2), ($\frac 13$, 3), ($\frac 14$, 4)}Ans

Here,

1. {(3,2), (-1,-7), (4,5), (6,8)}

Let f = {(3,2), (-1,-7), (4,5), (6,8)}

Changing domain into range and range into domain.

f-1 = {(2,3), (-7,-1), (5,4), (8,6)}

1. {x, 2x+1 : x∈ R}

Let y = f(x) = 2x+1

Interchange the place of x and y

x = 2y + 1

x-1 = 2y

2y = x-1

y = $\frac {x-1}{2}$

∴ f-1 = $\frac{x-1}{2}$ Ans

f(4x-15) = 8x-27

Or,

f(4x-15) = 2(4x-15) + 3

∴ f(x) = 2x+3

ff(x) = f(2x+3) = 2 (2x+3) +3 = 4x +6+3 = 4x+9

ff(2) = 4 × 2 + 9 = 17 Ans

g(2x-5) = 8x-19

g(2x-5) = 4(2x-5) +1

g(x) = 4x+1

gg(x) =g(4x+1) = 4(4x+1) +1 = 16x+4+1 = 16x+5

gg(2) = 16×2+5 = 32+5 =37 Ans

f(x+3) = 3x+5

f(x+3) = 3(x+3) -4

f(x) = 3x-4

Let y = f(x) = 3x-4

Y = 3x-4

Interchanging the place of x and y in the above relation

x = 3y-4

Or, 3y = x+4

y = $\frac {x+4}{3}$

f-1 (x) = $\frac {x+4}{3}$ Ans

fog(x) = 6x-2

Or, f[g(x)] = 6x-2

Or, f(2x) = 6x-2

Or, f(2⋅$\frac x2$ = 6 $\frac x2$-2

f(x) = 3x-2 Ans

Let g(x) = ax+b

fog(x) = f[g(x)] = f(ax+b) = 8(ax+b) -3

fog(x) = 8ax +8b - 3 ------ (1)

fog(x) = 20x+1 ------------(2)

From eqn (1) and (2)

8ax +8b -3 = 20x +1

Equating on both sides,

8a = 20

a = $\frac {20}{8}$

∴ a = $\frac 52$

8b -3 = 1

8b = 4

b =$\frac 48$ = $\frac 12$

∴ b = $\frac 12$

Putting the value of a and b in g(x) = ax + b

g(x) = $\frac 52$ x + $\frac 12$ = $\frac {5x+1}{2}$Ans

Function f = {(x, 2x+1) : x∈R}

Let, y = 2x + 1

Interchanging the place of x and y

x = 2y + 1

or, x-1 = 2y

or, y = $\frac {x-1}{2}$

∴ f-1= {(x, $\frac {x-1}2$) :x∈R } Ans

f(x) = 2x + 3, g (x) = 5x2

fg (x)

= f (5x2) [$\because$ g(x) = 5x2]

= 2×5x2+ 3

= 10x2 +3 Ans

Also,

gf(x)

= g (2x + 3) [$\because$ f(x) = 2x + 3 ]

= 5 (2x + 3)2Ans

fog (2)

= f(2x - 3) [$\because$ g(x) = x - 3 ]

= f(-1)

= (-1)2 + 5 [$\because$ f(x) = x2 + 5 ]

= 1 + 5

= 6 Ans

x3 + 3x2 - 2x + 6÷ (x - 3)

 3 1 3 -2 6 3 18 48 1 6 16 54

The remainder (R) = 54 Ans

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

Proof: Let, Q(x) be the quotient and R be the remainder when f(x) is divided by (x - a) then, f(x) = (x -a)⋅Q(x) + R.

When x = a, f(a) = (a -a)⋅Q(x) + R ∴ f(a) = R proved

If the polynomial f(x) is divided by (x - a) and R = f(a) = 0, then (x - a) is a factor of f(x).

Proof: Let Q(x)

Let Q(x) be the quotient when f(x) is divided by (x - a), then f(x) = (x-a) Q(x) + R = (x-a) Q(x) + 0

f(x) = (x-a) Q(x)

∴(x-a) is a factor of f(x). proved

x4 - x3 - 3x2 -2x + 5÷ (x + 1)

 -1 1 -1 -3 -2 5 -1 2 1 1 1 -2 -1 -1 6

∴ Quotient (Q) = x3 - 2x2 - x -1

Remainder (R) = 6 Ans

x3 - 2x + 3 is divided by x - 1.

 1 1 0 -2 3 1 1 -1 1 1 -1 2

Quotient (Q) = x2 - x -1

Remainder (R) = 2 Ans

Here,

f(x)÷ g(x) = $\frac {x^3 +3x^2 -4x +2}{x +1}$

 x + 1 x3 + 3x2 - 4x +2 x2 + 2x - 6 x3 + x2 - - 2x2 - 4x + 2 2x2 + 2x - - -6x + 2 -6x - 6 + + 8

∴ Quotient Q(x) = x2 + 2x - 6

Remainder R(x) = 8 Ans

Here,

x - k is factor of the polynomialx3 - kx2 - 2x + k + 4, then x = k must satisfies the given polynomial, so that the given polynomial should be zero.

x - k is a factor of f(k) then f(k) = 0

f(k) = k3 - k⋅ k2 - 2k + k + 4

or, 0 = k3 - k3 -k +4

or, 0 = -k +4

∴ k = 4 Ans

Here,

f(x) =2x3 + ax2 + x + 2

2x + 1 is a factor of f(x) so f(x) = 0

2x + 1 = o

or, 2x = -1

∴ x = - \frac 12

2x3 + ax2 + x + 2 = 0

or, 2× (- $\frac 12)^3$ + a⋅ (- $\frac 12)^2$ - $\frac 12$ + 2 = 0

or, 2× - $\frac 18$+ a⋅ $\frac 14$ - $\frac 12$ + 2 = 0

or, - $\frac 14$ + $\frac a4$ - $\frac 12$ + $\frac 21$ = 0

or, $\frac {-1 +a -2 +8}{4}$ = 0

or, a + 5 = 0

∴ a = -5 Ans

Arrange the given polynomial in decreasing order

f(x) = 4x3 -6x2+3x -5

The constant term with the sign changed is +2

∴ x - 2 = x - (+2)

Writes the coefficient of f(x) in decreasing order:

 2 4 -6 +3 -5 ↓ 8 4 14 4 2 7 9 Q R

∴ Quotient Q(x) = 4x2 +2x +7

Remainder R(x) = 9 Ans

Arranging the given polynomial in descending order

f(t) = 7t4 -4t3 +6t +3

The constant term with sign changed is + 3,

t - 3 = t - (+3)

Write the coefficient in descending order;

 +3 7 -4 0 6 3 ↓ 21 51 153 477 7 17 51 159 480 Q R

∴ Quotient Q(t) = 7t3 + 17x2 +51x +159

Remainder R(t) = 480 Ans

(x + 2) is a factor of x3 + kx2 - 4x + 12 or x + 2 = 0 or x = -2

Putting the value x = -2 in the given expression and f(-2) = 0

x3 + kx2 - 4x + 12 =0

or, (-2)3 + k(-2)2 - 4× (-2) + 12 = 0

or, -8 + 4k + 8 + 12 = 0

or, 4k = -12

or, k = - $\frac {12}{4}$ = -3

∴ k = -3 Ans

x - 3 is a factor of the given expression.

So, x=3 must satisfy the given expression, then f(3) = 0

f(x) = x3 + 4x2 + kx - 30

or, f(3) = 33 + 4×32 + k⋅3 - 30

or, 0 = 27 + 36 +3k -30

or, 3k + 33 = 0

or, 3k = -33

or k = -$\frac {33}{3}$ = -11

∴ k = -11 Ans

f(x) = x3 - kx2 - x - 2

If (x - 2) is a factor of f (x) then

f(2) = 0

f(2) =23 - k⋅22 - 2- 2

or, 0 = 8 - 4k - 4

or, 0 = 4 - 4k

or, 4k = 4

or, k = $\frac 44$

∴ k = 1 Ans

If f(a0 = 0, the remainder when f(x) is divided by (x - a) is zero then (x - a) is factor of f(x).

x + 1 = 0

or, x = -1

∴ f(-1) = 0

2x3 - kx2 - 8x + 5

Putting the value of x = -1 in above relation,

f(-1) = 2 (-1)3 - k (-1)2 - 8 (-1) + 5

or, - 2 - k + 8 + 5 = 0

or, 11 - k = 0

∴ k = 11 Ans

Quotient Q(x) =x2 + 2x + 1, Remainder R(x) = 2, Original polynomial = f(x)

We know,

f(x) = (x + 1) Q(x) + R(x)

or, f(x) = (x + 1) (x2 + 2x + 1) + 2

or, f(x) = x3 + 2x2 + x -x2 - 2x - 1 + 2

or, f(x) =x3 + x2- x + 1 Ans

Let,

f(k) =x3 - kx2 - 2x + k + 4

If x - k is a factor of f(k) then f(k) = 0

f(k) = k3 - k⋅k2 - 2k + k + 4

or, 0 = k3 - k3 - k + 4

or, 0 = -k + 4

∴ k = 4 Ans

x + 3 is a factor of the given expression.

When x = -3, putting the value in the given expression equal to zero.

(-3)3 - (k - 1) (-3)2 + k(-3) +54 = 0

or, -27 -9k + 9 - 3k + 54 = 0

or, -12k + 36 = 0

or, -12k = -36

or k = $\frac {-36}{-12}$

∴ k = 3 Ans

If a number C is substituted for x in the polynomial p(x) of degree n, then P(C) is the remainder that would be obtained by dividing p(x) by x - c.

i.e. P(x) = Q(x)⋅ (x - c) + P(c)

where, Q(x) is a polynomial of degree n -1

f(x) = 3x3 - 5x2 + 2x - 3 and g(x) = x - 2 = x - (-2)

If f(x)÷ g(x), quotient = Q and remainder (R) = ?

f(2) = 3(2)3 - 5(2)2 + 2×2 - 3 = 24 - 20 + 4 -3 = 28 - 23 = 5

Remainder (R) = 5 Ans

Remainder theorem: The remainder theorem states that if f(x) is divided by (x - a), then f(a) will be the remainder.

x + 1 is a factor ofx4 - 3x3 -2x2 +x +5

f(-1) =(-1)4 - 3(-1)3 -2(-1)2 +(-1) +5 = 1 +3 -2 -1 +5 = 6 Ans

x + 2 is a factor of 3x2 + px2 - 2x - 8, so the given expression is satisfies by x = -2

3x2 + px2 - 2x - 8 = 0

or, 3(-2)2 + p(-2)2 - 2(-2) - 8 = 0

or, -24 + 4p +4 -8 = 0

or, 4p = 28

∴ p = $\frac {28}{4}$ = 7 Ans

Factor Theorem: If a polynomial f(x) is divided by (x - a) and remainder R = f(a) = 0 then x - a is a factor of f(x). This theorem is known as the factor theorem.

Given,

f(x) = 2x4 - 3x2 + 6x + k and f(1) = 0

If x = 1 then

2 × 14 - 3 × 12 + 6 × 1 + k = f(1)

or, 2 - 3 + 6 + k = 0

or, 5 + k = 0

∴ k = -5 Ans

Constant Function: A function f:A → B is called a constant function if there exists an element C∈B such that f(x) = C for all X∈A. A set containing only one element.

Let: Y = f(x) = 7x - 8

Y = 7x -8 [∴ Range = 13]

or, 7x = 13 + 8

or, x = $\frac {21}{7}$ = 3

∴ Domain = 3 Ans

Remainder Theorem: When a polynomial f(x) is divided by a linear polynomial x - a then the remainder R is given by the value f(a) of the polynomial, R = f(a).

 x - 3 2x3 - 7x2 + 5x + 4 2x2 - x + 2 2x3 - 6x2 - 445 - + - - x2 + 5x + 4 - x2 + 3x + - 2x + 4 2x - 6 - + 10

∴ Remainder = 10 Ans

Remainder Theorem: If f(x) is a polynomial of degree n in x and if f(x)is divided by x - a, then the remainder is f(a). This theory is known as Remainder Theorey.

Let: f(x) =a4 - 3a3 - 2a2 + a + p

If a + 1 is factor of f(x) then f(-1) = 0

f(-1) =(-1)4 - 3(-1)3 - 2(-1)2 + (-1) + p

or, 0 = 1 + 3 - 2 - 1 + p

or, p + 1 = 0

∴ p = -1 Ans

f(2) = 8

f(2) =2×23 + 3×22 + k (1 - $\frac {3×2}{k}$ )

or, 8 = 16 + 12 +k (1- $\frac 6k$)

or, 28 + k $\frac {k-6}{k}$ = 8

or, k - 6 = 8 - 28

or, k = -20 + 6

∴ k = -14 Ans

f(x) =3x3 - 2x2 + 4x - 1 and g(x) = 3x + 2

g(x) = x + $\frac 23$ = x - (- $\frac 23$)

In f(x)÷ g(x), remainder R = f(a) where a = - $\frac 23$

f(-$\frac 23$) = 3 (-$\frac 23)^3$ + 2 (-$\frac 23)^2$ + 4×(-$\frac 23$) - 1

= -$\frac 89$ + $\frac 89$ - $\frac 83$- 1

= $\frac {-8-3}{3}$

= -$\frac {11}{3}$

∴ Remainder R =f(- $\frac 23$) =-$\frac {11}{3}$ Ans

f(x) =6x3 - (k + 6)x2 + 2kx - 25

If (2x - 5) is a factor of f(x) then f($\frac 52$) = 0

f(\frac 52) = 6($\frac 52)^3$ - (k + 6)($\frac 52)^2$ + 2k($\frac 52$) - 25

or, 0 = $\frac {375}{4}$ - $\frac{25}{4}$ (k + 6) + 5k - 25

or, 0 = $\frac {375 - 25k - 150 + 20k - 100}{4}$

or, -5k + 125 = 0

or, -5k = -125

or, k = $\frac {-125}{-5}$

∴k = 25 Ans

Let, f: A→B be a one to one onto function then a function f-1 : B→A is called an inverse function of f. i.e.

f = {(2, 5), (3, 6), (-4, 1). (7, 4)}

f-1 ={(5, 2), (6, 3), (1, -4). (4, 7)} Ans

Here,

f(x) = 2x3 + 3x2 - 3x + p and f(2) = 8

Putting the value of x = 2,

f(2) = 2 × 23 + 3 × 22 - 3 × 2 + p

or, 8 = 16 + 12 - 6 + p

or, p = 8 - 22

∴ p = -14 Ans

Here,

f(x) = 2x4 - 3x2 + 6x + k and f(1) = 0

Putting the value of x = 1

f(1) = 2 × 14 - 3 × 12 + 6 × 1 + k

or, 0 = 2 -3 + 6 + k

or, k + 5 = 0

∴ k = - 5 Ans

Here,

f(x) = x2 and g(x) = 3x

1. fog(x) = f(3x) = (3x)2 =9x2Ans
2. gof(x) = g(x2) = 3x2Ans
3. fog(2) = 9x2 = 9 (2)2 = 36 Ans
4. gof(2) = 3x2 = 3 (2)2 = 12 Ans

i. gof(x)

= g(2x - 3) [$\because$ f(x) = 2x - 3]

= 3(2x - 3) + 4 [$\because$ g(x) = 3x + 4]

= 6x - 9 + 4

= 6x -5 Ans

Let, y = f(x) = 2x - 3

∴ y = 2x - 3

Interchanging the place of x and y

x = 2y - 3

or, 2y = x + 3

or, y = $\frac {x+3}{2}$

i.e. f-1(x) =$\frac {x+3}{2}$

Let, y= g(x) = 3x + 4

∴ y = 3x + 4

Interchanging the place of x and y

x = 3y + 4

or, 3y = x - 4

or, y = $\frac {x-4}{3}$

i.e. g-1(x) =$\frac {x-4}{3}$

ii. fog-1(x)

= f$\frac {x-4}{3}$ [$\because$ g-1(x) =$\frac {x-4}{3}$ ]

= 2 $\frac {x-4}{3}$ -3

= $\frac {2x- 8 - 9}{3}$

= $\frac {2x - 17}{3}$Ans

iii. f-1og (2)

= f-1(3× 2 + 4)

=f-1(10)

= $\frac {10 + 3}{2}$ [$\because$ $\frac {x+3}{2}$]

= $\frac {13}{2}$ Ans

iv. fog(-3)

= f(3× -3 + 4) [$\because$ g(x) = 3x + 4]

= f( -9 + 4)

= f(-5)

= 2× (-5) - 3

= -10 - 3

= -13 Ans

let, y= f(x) = 8 - 3x

Interchanging the place of x and y

x = 8 - 3y

or, 3y = 8 - x

∴ y = $\frac {8 - x}{3}$

i.e. f-1(x) = $\frac {8 - x}{3}$

i. f-1(-4) = $\frac {8 - (-4)}{3}$ = $\frac {12}{3}$ = 4 Ans

ff(x) = f(8 - 3x) = 8 - 3(8- 3x) = 8 - 24 + 9x = 9x - 16

ff(2) = 9× 2 - 16 = 18 - 16 = 2 Ans

Let, y = g(x) = 3x - 5

y = 3x - 5 ---------- (1)

Interchanging the place of xand y in equation (1)

x = 3y - 5

or, 3y = x + 5

∴ y = $\frac {x + 5}{3}$

i.e. g-1(x) = $\frac {x + 5}{3}$

fg-1(x) = 15

or, f$\frac {x + 5}{3}$=15

or, 4$\frac {x + 5}{3}$ = 15

or, $\frac {4x + 20 + 21}{3}$ = 15

or, 4x + 41 = 45

or, 4x = 45 - 41 = 4

or, x = $\frac 44$

∴ x = 1 Ans

Let, y = f(x) = 3x + 5

or, y= 3x + 5

Interchanging the position of x and y,

x = 3y + 5

or, 3y = x - 5

or, y = $\frac {x - 5}{3}$

∴ f-1(x) =$\frac {x - 5}{3}$

f-1(2) =$\frac {2 - 5}{3}$ = - $\frac 33$ = -1

ff-1(2) = f(-1) = 3× (-1) + 5 = - 3 + 5 = 2

f-1f(3) = f-1 (3× 3 +5) = f-1 (14) = $\frac {14 - 5}{3}$ = $\frac 93$ = 3

Hence, ff-1(2) = 2

f-1 f(3) = 3 Ans

ghf(x)

= gh(3x - 4)

= g [ -2 (3x - 4) + 1]

= g(-6x + 8 + 1)

= g(- 6x + 9)

= -6x + 9 +3

= - 6x + 12

Let,

y = g(x) = x + 3

∴ y = x + 3

Interchanging the place of x and y,

x = y + 3

or, x - 3 = y

i.e y = x - 3 [$\because$ g-1(x) = x - 3]

g-1hf(x)

= g-1h(3x - 4)

= g-1 [ - 2(3x - 4) + 1]

= g-1 (-6x + 8 + 1)

= g-1 (-6x + 9)

= - 6x + 9 -3

= - 6x + 6

Hence, ghf(x) = - 6x + 12

and, g-1hf(x) = -6x + 6 Ans

Given,

f(x) = 3x + 4

g(x) = 2(x + 1) = 2x + 2

fog(x) = f(2x + 2) = 3 (2x + 2) + 4 = 6x + 6 + 4 = 6x + 10

gof(x) = g(3x + 4) = 2 (3x + 4) + 2 = 6x + 8 +2 =6x + 10

Hence, fog(x) = gof(x) proved

Let y = 3x + 4 = f(x)

y = 3x + 4

Interchanging the place of x and y

x = 3y + 4

or, 3y = x - 4

or, y = $\frac {x - 4}{3}$

f-1(x) = $\frac {x - 4}{3}$

∴ f-1(2) = $\frac {2 - 4}{3}$ = - $\frac 23$Ans

Let y = f(x) = $\frac 1x$

y = $\frac 1x$

Interchanging the place of x and y

x = $\frac 1y$

y = $\frac 1x$

∴ f-1(x) = $\frac 1x$

L.H.S

=fof-1 (x)

= f$\frac 1x$

=$\cfrac{1}{ \cfrac{1}{x}}$

= x

R.H.S

= f-1of(x)

= f-1$\frac 1x$

= $\cfrac{1}{ \cfrac{1}{x}}$

= x

Hence, L.H.S = R.H.S proved

Let y = f(x) =$\frac {x}{x-3}$

y = $\frac {x}{x - 3}$

Interchanging the place of x and y,

x = $\frac {y}{y - 3}$

or, xy- 3x= y

or, xy - y = 3x

or, y (x - 1) = 3x

or, y = $\frac {3x}{x - 1}$

i.e., f-1(x) = $\frac {3x}{x - 1}$

From the question, f(x) = f-1(x)

$\frac {x}{x - 3}$ =$\frac {3x}{x - 1}$

or, x2 - x = 3x2 - 9x

or, 3x2 - 9x - x2 + x = 0

or, 2x2 - 8x = 0

or, 2x(x - 4) = 0

Either, x = 0 and x - 4 = 0 or, x = 4

∴x = 0, 4 Ans

Here,

f(x) = x2 - 2x and g(x) = 2x + 3

Let: y = g(x) = 2x + 3

Interchanging the place of x and y

x = 2y + 3

or, 2y = x - 3

or, y = $\frac {x-3}{2}$

∴ g-1(x) =$\frac {x-3}{2}$

fg-1(x) = 3

f$\frac {x-3}{2}$ = 3

or, $(\frac {x-3}{2})^2$ - 2 $\frac {x-3}{2}$ = 3

or, $\frac {x^2 - 6x + 9}{4}$ -$\frac {x-3}{2}$ = 3

or, $\frac {x^2 - 6x + 9 - 4x + 12}{4}$ = 3

or, x2 - 10x + 21 =12

or, x2 -10x + 21 - 12 = 0

or, x2 - 10x + 9 = 0

or, x2 - 9x - x + 9 = 0

or, x(x - 9) -1 (x - 9)= 0

or, (x - 9) (x - 1) = 0

Either, x - 9 = 0 i.e. x = 9

or, x - 1 = 0 i.e. x = 1

∴ x = 9 and 1 Ans

Let: y =g(x) = $\frac {1}{1-x}$

y = $\frac {1}{1-x}$

Interchanging the place of x and y

x =$\frac {1}{1-y}$

or, 1 - y = $\frac 1x$

or, y = 1 - $\frac 1x$ = $\frac {x - 1}{x}$

g-1(x) = $\frac {x - 1}{x}$

g-1$\frac 12$

= $\cfrac{(\frac{1}{2})-1}{ \cfrac{1}{2}}$

= $\frac {1-2}{2}$× $\frac 21$

= -1 Ans

Again, fg(x) = f$\frac {1}{1-x}$ = 1 + $\frac {2×1}{1-x}$

fg(x) = f$\frac {1}{1-x}$ = 1 + $\frac {2×1}{1-x}$

= f$\frac {1}{1-x}$ = 1 + $\frac {2×1}{1-x}$

= 1 + $\frac {2×1}{1-x}$

= $\frac {1 - x + 2}{1 - x}$

= $\frac {3 - x}{1 - x}$

fg(-1)

= $\frac {3 - (-1)}{1 - (-1)}$

= $\frac {3 + 1}{1 + 1}$

= $\frac 42$

= 2 Ans

Let: y = f(x) = $\frac {2x + 3}{x + 2}$

Interchanging the position of x and y

x= $\frac {2y + 3}{y + 2}$

or, xy+ 2x - 2y = 3

or, y(x - 2) = 3 - 2x

∴y= $\frac {3 - 2x}{x - 2}$

1. f-1(x) = $\frac {3 - 2x}{x - 2}$
2. f-1(1) = $\frac {3 - 2(1)}{1 - 2}$ = -$\frac 11$ = -1
3. fg(x) = f(x - 2) =$\frac {2(x-2) + 3}{(x-2) + 2}$ =$\frac {2x - 4 + 3}{x}$ = $\frac {2x-1}{x}$
4. fg(1) = $\frac {2×1-1}{1}$ = $\frac {2-1}{1}$Ans

Let, y= f(x) =$\frac {3x + 11}{x - 3}$

Interchanging the value of x and y,

x =$\frac {3y + 11}{y - 3}$

or, xy - 3x = 3y + 11

or, xy - 3y = 3x + 11

or, y(x - 3) = 3x + 11

or, y = $\frac {3x + 11}{x - 3}$

∴ f-1(x) = $\frac {3x + 11}{x - 3}$ Ans

Again,

Let: y = g(x) = $\frac {x - 3}{2}$

y =$\frac {x - 3}{2}$

Interchanging the place of x and y,

x =$\frac {y - 3}{2}$

or, 2x = y - 3

or, y = 2x + 3

∴ g-1(x) = 2x + 3

From the given question,

f(x) = g-1(x)

or,$\frac {3x + 11}{x - 3}$ = 2x + 3

or, 3x + 11 = 2x2 + 3x - 6x - 9

or, 2x2 - 3x - 9 - 3x - 11 = 0

or, 2x2 - 6x - 20 = 0

or, 2(x2 - 3x - 10) = 0

or, x2 - 5x + 2x - 10 = 0

or, x(x - 5) + 2(x - 5) = 0

or, (x - 5) (x + 2) = 0

Either, x - 5 = 0∴ x = 5

Or, x + 2 = 0∴ x = -2

Hence, x = 5 or -2 Ans

Here,

Let: y = g(x) = 3x - 5

y = 3x - 5 ------------------(1)

Interchanging the position of x and y in equation (1),

x = 3y - 5

or, 3y = x + 5

or, y = $\frac {x + 5}{3}$

∴ g-1(x) = $\frac {x + 5}{3}$

fg-1(x) = 15

or, f $\frac {x + 5}{3}$ = 15

or, 4 $\frac {x + 5}{3}$ + 7 = 15

or, $\frac {4x + 20}{3}$ + 7 = 15

or, $\frac {4x + 20 + 21}{3}$ = 15

or, 4x + 41 = 45

or, 4x = 45 - 41

or, 4x = 4

or, x = $\frac 44$

∴ x = 1 Ans

x3 - 4x2 + x + 6 = 0

(x - 2) is a factor ofx3 - 4x2 + x + 6

or, x3 - 2x2 - 2x2 + 4x - 3x + 6 = 0

or, x2(x - 2) - 2x(x - 2) - 3(x - 2) = 0

or, (x - 2) (x2 - 2x - 3) = 0

or, (x - 2) (x2 - 3x + x - 3) = 0

or, (x - 2) [x(x - 3) + 1(x - 3)] = 0

or, (x - 2) (x - 3) (x + 1) = 0

Either, x - 2 = 0 ∴ x = 2

Or, x - 3 = 0 ∴ x = 3

or, x + 1 = 0 ∴ x = - 1

Hence, x = 2, 3, -1 Ans

Rough:

x = 2

x3 - 4x2 + x + 6

= 23 - 4⋅ 22 + 2 +6

= 16 - 16

= 0

x - 2 is a factor of given expression

or, 2x3 - 4x2 + 7x2 - 14x + 3x - 6 = 0

or, 2x2 (x - 2) + 7x (x - 2) + 3 (x - 2) = 0

or, (x - 2) (2x2 + 7x + 3) = 0

or, (x - 2) (2x2 + 6x + x +3) = 0

or, (x - 2) [2x (x + 3) + 1 (x + 3)] = 0

or, (x - 2) (x + 3) (2x + 1) = 0

Either, x + 3 = 0∴ x = -3

Or, x - 2 = 0∴ x = 2

Or, 2x + 1 = 0∴ x = -$\frac 12$

∴ x = 2 , - 3 , -$\frac12$ Ans

Rough:

If x = 2

2x3 + 3x2 - 11x - 6

= 2(2)3 + 3(2)2 - 11(2) - 6

= 16 + 12 - 22 - 6

= 0

x = -1 is correct.

x + 1 is a factor of above equation.

or, 3x3 + 3x2 - 16x2 - 16x + 16x + 16 = 0

or, 3x2(x + 1) - 16x(x + 1) + 16(x + 1) = 0

or, (x + 1) (3x2 - 16x + 16) = 0

or, (x + 1) (3x2 - 12x - 4x + 16) = 0

or, (x + 1) [3x (x - 4) - 4 (x - 4)] = 0

or, (x + 1) (x - 4) (3x - 4) = 0

Either, x + 1 = 0 ∴ x = - 1

Or, x - 4 = 0 ∴ x = 4

Or, 3x - 4 = 0 ∴ x = $\frac 43$

∴ x = -1, 4 and $\frac 43$ Ans

Rough:

Put x = -1

or, 3(-1)3 - 13(-1)2 + 16 = 0

or, -3 -13 + 16 = 0

or, - 16 + 16 = 0

∴ 0 = 0

x + 2 is a factor of given equation.

x3 + 2x2 - 2x2 - 4x - 15x - 30 = 0

or, x2(x + 2) - 2x(x + 2) - 15(x + 2) = 0

or, (x + 2) (x2 - 2x - 15) = 0

or, (x + 2) (x2 - 5x + 3x - 15) = 0

or, (x + 2) [x(x - 5) + 3(x - 5)] = 0

or, (x + 2) (x - 5) (x + 3) = 0

Either, x + 2 = 0 ∴ x = -2

Or, x - 5 = 0 ∴ x = 5

Or, x + 3 = 0 ∴ x = -3

∴ x = -2, -3, 5 Ans

x3 - 3x - 2 = 0

x + 1 is a factor of given equation,

or, x3 + x2 - x2 - x - 2x - 2 = 0

or, x2 (x + 1) - x (x + 1) - 2 (x + 1) = 0

or, (x + 1) (x2 - x - 2) = 0

or, (x + 1) [x(x - 2) + 1 -(x - 2)] = 0

or, (x + 1) (x - 2) (x + 1) = 0

Either, x + 1 = 0 ∴ x = -1

Or, x - 2 = 0 ∴ x = 2

∴ x = -1 and 2 Ans

(x + 1) (x + 2) (x + 3) (x + 4) - 8 = 0

or, (x2 + x + 4x + 4) ( x2 + 2x + 3x + 6) - 8 = 0

or, (x2 + 5x + 4) (x2 + 5x + 6) - 8 = 0

Let: x2 + 5x = k

(k + 4) (k + 6) - 8 = 0

or, k2 + 6k + 4k + 24 - 8 = 0

or, k2 + 10k + 16 = 0

or, k2 + 8k + 2k + 16 = 0

or, k(x + 8) + 2(x + 8) = 0

or, (k + 2) (x + 8) = 0

Putting the value of k

(x2 + 5x + 8) (x2 + 5x + 2) = 0

Either, x2 + 5x + 8 = 0 --------------(1)

Or, x2 + 5x + 2 = 0 ------------------(2)

Taking eqn(1)

x =$\frac {-5 ± (\sqrt{5^{2}-4×1×8})}{2×1}$ =$\frac {-5 ± \sqrt{-7}}{2×1}$ (Impossible)

Taking eqn(2)

x =$\frac {-5 ± (\sqrt{5^{2}-4×1×2})}{2×1}$ =$\frac {-5 ± \sqrt{17}}{2×1}$ Ans

Here,

f(x) = x4 - x3 - 3x2 - 2x + 5

When x = 2 then,

 2 1 -1 -3 -2 5 2 2 -2 -8 1 1 -1 -4 -3

∴ Quotient = x3 + x2 - x - 4 and f(x) = -3 Ans

When x = -1 then,

 -1 1 -1 -3 -2 5 -1 2 1 1 1 -2 -1 -1 6

∴ Quotient = x3 - 2x2 - x - 1 and f(2) = 6 Ans

When x = 3 then,

 3 1 -1 -3 -2 5 3 6 9 21 1 2 3 7 26

∴ Quotient = x3 + 2x2 + 3x + 7 and f(3) = 26 Ans

x - 2 = 0

x - 2 is a factor of given expression

or, x3 - x2 - 14x + 24 = 0

or, x3 - 2x2 + x2 - 2x - 12x + 24 = 0

or, x2(x - 2) + x(x- 2) - 12(x - 2) = 0

or, (x - 2) (x2 + x -12) = 0

or, (x - 2) (x2 + 4x - 3x - 12) = 0

or, (x - 2) [x(x + 4) - 3(x + 4)] = 0

or, (x - 2) (x + 4) (x -3) = 0

Either, x - 2 = 0 ∴x = 2

Or, x + 4 = 0 ∴ x = - 4

Or, x - 3 = 0 ∴ x = 3

∴ x = 2, 3, -4 Ans

Rough:

x = 2

i.e 23 - 22 - 14× 2 + 24

= 8 - 4 -28 + 24

= 0

2x3 - 3x2 - 3x + 2 = 0

or, 2x3 + 2 - 3x2 - 3x = 0

or, 2(x3 + 13) - 3x (x + 1) = 0

or, 2(x + 1) (x2 - x + 1) - 3x (x + 1) = 0

or, (x + 1)[2x2 - 2x + 2 - 3x] = 0

or, (x + 1) (2x2 - 5x + 2) = 0

or, (x + 1) (2x2 - 4x - x + 2) = 0

or, (x + 1) [2x(x - 2) - 1(x - 2)] = 0

or, (x + 1) (x - 2) (2x - 1) = 0

Either, x + 1 = 0 ∴x = -1

Or, x - 2 = 0 ∴x = 2

Or, 2x - 1 = 0 ∴ x = $\frac 12$

∴ x = -1, 2 , $\frac12$ Ans