## Algebraic Expression for SBI PO

# Algebraic Expression for SBI PO

**Algebraic Expression : **Algebraic expression for SBI PO is an expression that contains variables , constants and algebraic operations (addition, subtraction, multiplication, division)

**Example** x^{2}+ 3y = 7

2xy = 61

**Polynomial:** An algebraic expression for SBI PO which can be written in the form

p(x) = a_{0}+ a_{1} x + a_{2 }x^{2 }+ ………………+ a_{n} x^{n} is called a polynomial in x, where a_{0},a_{1}, ….., a_{n} belongs to real numbers and n is a non – negative integer.

I.e. an algebraic expression which contains only non -negative powers of x is called a polynomial in x.

**Example: **2x^{3 }– 4x + 5

x^{4} + 2x^{2 }– x + 2

Theory for Algebraic Expression for SBI PO

**Degree of a polynomial:** The highest power of x present in a polynomial is called the degree of a polynomial.

**Example:** 2x^{3 }– 4x + 5 _The highest power of x in this polynomial is 3, so its degree is 3

x^{4}+ 2x^{2}– x + 2 _ the highest power of x in this polynomial is 4, so its degree is 4

**Constant polynomial:** A polynomial of degree 0 is called a constant polynomial

Example: 4

56

-8/5

**Zero polynomial:** p(x) = 0 is called a zero polynomial. The degree of zero polynomial is not defined

**Linear polynomial:** A polynomial of degree 1 is called a linear polynomial.

**Example:** x + 3

2x – 5

**Quadratic polynomial:** A polynomial of degree 2 is called a quadratic polynomial.

**Example:** 4x^{2 }+ x + 6

x^{2 }– 4

**Following Question Prepared by Best Bank Coaching in Noida**

**Cubic polynomial:** A polynomial of degree 3 is called a cubic polynomial.

**Example:** x^{3} – 4x^{2 }+ x + 6

10x^{3}+ 85

**Remainder theorem:** Let p(x) be any polynomial and let a be any real number . if p(x) is divided by the linear polynomial x – a , then the remainder is p(a).

In general when p(x) is divided by the linear polynomial ax + b , the remainder is p(-b/a).

**Example:** if we divide, p(x) = x^{3}– 4 x^{2}+ x + 6 by 2x + 4 the remainder is p(-4/2) = p(-2)

p(-2) = (-2)^{3}– 4(-2^{2}) + (-2) + 6

= – 8 – 16 – 2 + 6

= -20

**Factor theorem:** If p(x) is a polynomial of degree n ≥ 1 and a is any real number, then x – a is a factor of p(x), if (if and only if ) p(a) = 0

**Top BANK Coaching Center in New Delhi**

**Example:** Check whether x + 2 is a factor of p(x) = x^{2 }+ 4x + 4

We first put x + 2 = 0

x = -2

Now we calculate p(-2)

p(-2) = (-2)^{2}+ 4(-2) + 4

= 4 – 8 + 4 = 0

p(-2) = 0

⇒ x + 2 is a factor of x^{2}+ 4x + 4

**Greatest integer:** The greatest integer of a real number x is the greatest integer less than or equal to the x. It is denoted by x

**Example:** [4.72] = 4

[62] = 62

[-1.23] = -2

**Minimum value of a quadratic polynomial:** Minimum value of a quadratic polynomial ax^{2 }+ bx + c is given by the formula

4ac – b^{2}/4a

** **** Algebraic Expression for SBI PO**

**(a + b )**^{2}= a^{2}+ b^{2}+ 2ab- (a – b )
^{2}= a^{2}+ b^{2}– 2ab - a
^{2}– b^{2}= (a + b) (a – b) - (a + b )
^{3 }= a^{3}+ b^{3 }+ 3ab(a + b) - (a – b )
^{3 }= a^{3}– b^{3 }– 3ab(a – b) - a
^{3}+ b^{3 }= (a + b) ( a^{2}+ b^{2}– ab) - a
^{3}– b^{3 }= (a – b) ( a^{2}+ b^{2}+ ab) - a
^{3}+ b^{3 }+ c^{3}– 3abc = (a + b + c) (a^{2}+ b^{2}+ c^{2}– ab – bc – ca) - a
^{3}+ b^{3 }+ c^{3}– 3abc = 1/2 (a + b + c) [(a – b )^{2}+ (b – c )^{2}+ (c – a )^{2}] - (a + b + c) (a
^{2}+ b^{2}+ c^{2}– ab – bc – ca) = 1/2 (a + b + c) [(a – b )^{2}+ (b – c )2+ (c – a )^{2}] - If a + b + c = 0, then a
^{3}+ b^{3 }+ c^{3}= 3abc - (a + b + c)3 = a
^{3}+ b^{3}+ c^{3}+ 3(a + b) (b + c) (c + a) - (a + b + c)2= a
^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2ca

**Factors of some polynomial**

- x
^{n}+ y^{n}is exactly divisible by (x + y) only when n is odd.

Example x^{5} + y^{5} is exactly divisible by x + y

- x
^{n}+ y^{n}is not exactly divisible by x + y when n is even - x
^{n}+ y^{n}is never divisible by x – y - x
^{n}– y^{n}is exactly divisible by x + y when n is even - x
^{n}+ y^{n}is exactly divisible by x – y

**Quadratic equations :** A general form of quadratic equation is ax2 + bx + c = 0, where a,b and c are real numbers and a 0

**Example** 3x^{2} + 5x – 8 = 0

**Roots or zeros of quadratic equations :** The values of x which satisfies the given equation are called its roots.

**Example** x = 4 and x = -3 are the roots of the equation x^{2} – x – 12 = 0

A quadratic equation has exactly two roots

**Join Best Exam Preparation group on Telegram **

**Methods of finding the roots of quadratic equations**

**Method of splitting the middle term**suppose we have to find the roots of the equation x^{2}– x – 12 = 0

Find the product of constant term and coefficient of x^{2}

-12*1 = -12

Now find two numbers whose product is -12 and sum is -1 (coefficient of x)

That is -4 and 3

x^{2} – x – 12 = x^{2} – 4x + 3x – 12 = 0

= x(x – 4) + 3(x – 4) = 0

= (x – 4) (x + 3) = 0

= x – 4 = 0 or x + 3 = 0

= x = 4 or x = -3

- By quadratic formula or sridharacharya formula : The quadratic equation ax2 + bx + c = 0 where a,b and c R and a 0 has two roots

X = -b + b^{2} – 4a c^{2 }a and x = -b – b^{2} – 4a c^{2 }a

**Best books for Competitive Exam**

Discriminant of a quadratic equations : b^{2} – 4ac is called the discriminant of the quadratic equation ax^{2} + bx + c = 0 where a,b and c R and a 0 and is denoted by D.

Nature of roots of the quadratic equation :

- When D < 0, both roots will be imaginary
- When D = 0, both roots will be real and equal
- When D > 0, both roots will be real and distinct

Consider a quadratic equation

Y = x^{2} + 2x – 1

Graph of this equation is given below

The graph cut the x axis between (-2 and -3) and between (0 and 1)

So the roots of this equation lies exactly where graph cut the x axis

-1 + 2 and -1 – 2

Consider the equation y = x^{2} – 5x + 6

The roots of this equation are 2 and 3

Because the graph cut the x axis at 2 and 3

### Related Posts

« Question for Algebraic Expression for SBI Exam Classification Puzzle Test for Bank Exam »