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Algebraic Expression for SBI PO

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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 x2+ 3y = 7

               2xy = 61

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

p(x) = a0+ a1 x + ax+ ………………+ an xn is called a polynomial in x, where a0,a1, ….., an 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– 4x + 5

                x4 + 2x– 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– 4x + 5 _The highest power of x in this polynomial is 3, so its degree is 3

x4+ 2x2– 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+ x + 6

                x– 4

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

Example: x3 – 4x+ x + 6

               10x3+ 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) = x3– 4 x2+ x + 6 by 2x + 4 the remainder is p(-4/2) = p(-2)

p(-2) = (-2)3– 4(-22) + (-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

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Example: Check whether x + 2 is a factor of p(x) = x+ 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 x2+ 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+ bx + c is given by the formula

                      4ac – b2/4a

                     Algebraic Expression for SBI PO

  1. (a + b )2 = a2 +  b2 + 2ab
  2. (a – b )2 = a2 +  b2 –  2ab
  3. a2 –  b2 = (a + b) (a – b)
  4. (a + b )= a3 + b+ 3ab(a + b)
  5. (a – b )= a3 – b– 3ab(a – b)
  6. a3 + b= (a + b) ( a2 +  b2 –  ab)
  7. a3 – b= (a – b) ( a2 +  b2 +  ab)
  8. a3 + b+ c3 – 3abc = (a + b + c) (a2+ b2 + c2– ab – bc – ca)
  9. a3 + b+ c3 – 3abc = 1/2 (a + b + c) [(a – b )2 + (b – c )2+ (c – a )2]
  10. (a + b + c) (a2+ b2 + c2– ab – bc – ca) = 1/2 (a + b + c) [(a – b )2 + (b – c )2+ (c – a )2]
  11. If a + b + c = 0, then a3 + b+ c3= 3abc
  12. (a + b + c)3 = a3 + b3+ c3+ 3(a + b) (b + c) (c + a)
  13. (a + b + c)2= a2 +  b2+ c2 + 2ab + 2bc + 2ca

Factors of some polynomial

  • xn + yn is exactly divisible by (x + y) only when n is odd.

       Example x5 + y5 is exactly divisible by x + y

  • xn + yn is not exactly divisible by x + y when n is even
  • xn + yn is never divisible by x – y
  • xn – yn is exactly divisible by x + y when n is even
  • xn + yn 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 3x2 + 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 x2 – x – 12 = 0

A quadratic equation has exactly two roots

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Methods of finding the roots of quadratic equations

  1. Method of splitting the middle term suppose we have to find the roots of the equation     x2 – x – 12 = 0

Find the product of constant term and coefficient of x2

-12*1 = -12

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

That is -4 and 3

x2 – x – 12  = x2 – 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

  1. 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 + b2 – 4a ca     and  x = -b – b2 – 4a ca

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Discriminant of a quadratic equations : b2 – 4ac is called the discriminant of the quadratic equation  ax2 + 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 = x2 + 2x – 1

Graph of this equation is given below

capture

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 = x2 – 5x + 6

capture

The roots of this equation are 2 and 3

Because the graph cut the x axis at 2 and 3

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