ALGEBRA - solved problems
  Algebraic expressions
Simplifying algebraic expressions
1.    Simplify algebraic expressions.
Solutions:   a)   - 4a3 + 3a2 + 5a3 - 7a2 = (- 4 + 5) · a3 + (3 - 7) · a2  = a3 - 4a2,
b)   (x2  - x + 1) · (x + 1) = x3 - x2 + x + x2  - x + 1 = x3 + 1.
Evaluating algebraic expressions
2.    Evaluate the expression  x2  - 6xy + 9y2  for x = 2  and  y =  -1.
Solution: x2  - 6xy + 9y2 = 22  - 6 · 2 · (- 1)  + 9 · (-1)2 = 4 + 12 + 9 = 25.
Expanding algebraic expression by removing parentheses ( brackets)
3.    Expand given expressions.
Solutions:   a)   (a - b)2 = (a - b) · (a - b) = a2 - ab - ab + b2 = a2 - 2ab + b2,
b)   (a - b) · (a + b) = a2 - ab + ab - b2 = a2 - b2,
c)   (x + y) · (x2 - xy + y2) = x3 - x2y + xy2 + x2y  - xy2 + y3 = x3 + y3.
The square of a binomial (or binomial square)
4.    Square given binomials.
Solutions:   a)   (a + b)2 = (a + b) · (a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2,
b)   (2x + 3)2 = (2x)2 + 2 · (2x) · 3 + 32 = 4x2 + 12x + 9,
c)   (x - 2y)2 = x2  + 2 · x · (-2y) + (-2y)2 = x2 - 4xy + 4y2.
Squaring trinomial (or trinomial square)
5.    Square given trinomials.
Solutions:   a)  (x2 - 2x + 5)2 = (x2)2 + (2x)2 + 52 + 2 · x2 · (-2x) + 2 · x2 · 5 + 2 · (-2x) · 5 =
                           = x4 + 4x2 + 25 - 4x3 + 10x2 - 20x = x4  - 4x3 + 14x2 - 20x + 25,
b)  (a3 - a2b - 3ab2)2 = (a3)2 + (a2b)2 + (3ab2)2 + 2a3 (-a2b) + 2a3 (-3ab2) + 2(-a2b) (-3ab2) =
    = a6 + a4b2 + 9a2b4 - 2a5b - 6a4b2 + 6a3b3  = a6 - 5a4b2 + 9a2b4 - 2a5b + 6a3b3.
Cube of a binomial
6.    Cube (rise to third power) given binomials.
Solutions:   a)  (a - b)3 = (a - b)2 · (a - b) = (a2 - 2ab + b2) · (a - b) = 
                  = a3 - 2a2b + ab2 - a2b + 2ab2 - b3 = a3 - 3a2b + 3ab2 - b3,
b)  (x - 2)3 = x3 + 3 · x2 · (-2) + 3 · x · (-2)2  + (-2)3 = x3  - 6x2 + 12x - 8,
c)  (2x + y)3 = (2x)3 + 3 · (2x)2 · y + 3 · (2x) · y2  + y3 = 8x3 + 12x2y + 6xy2  + y3.
  Factoring algebraic expressions
Factoring algebraic expression by finding (determining) a common factor
7.    Factorize given expressions.
Solutions:   a)  3x - 6y = 3 · (x - 2y),     b)  xy - y2  = y · (x - y),     c)  a - a2  = a · (1 - a),
d)  x3 -3x2  + x = x · (x2 - 3x +1),    e)  x(a + b) - (a + b) = (a + b) · (x - 1),
f)   a(x - 3y) - x + 3y  = a(x - 3y) - (x - 3y) = (x - 3y) · (a - 1).
Grouping like terms, grouping and factorizing four terms
8.    Factorize given expressions.
Solutions:   a)  ax - bx - a + b = x(a - b) - (a - b) = (a - b) · (x - 1),
b)  a - 1 - ab + b = (a - 1) - b · (a - 1) = (a - 1) · (1 - b),
c)  x2 + ax - bx - ab = x(x + a) - b · (x + a) = (x + a) · (x - b),
d)  5ab2 - 3a3 - 10b3 + 6a2b = 5b2(a - 2b) -3a2(a - 2b) = (a - 2b)(5b2 - 3a2).
The square of a binomial - perfect squares trinomials
9.    Factorize given expressions.
Solutions:   a)  1 - 4x + 4x2 = 12  - 2 · 2x + (2x)2 = (1 - 2x)2  = (1 - 2x) · (1 - 2x),
b)  a5 + 6a4b + 9a3b2 = a3 · (a2  + 6ab  + 9b2 ) = a3(a + 3b)2 = a3(a + 3b)(a + 3b).
Difference of two squares
10.    Factorize given expressions.
Solutions:   a)  16x2 - 1 = (4x)2 - 12 = (4x -1) · (4x +1),
b)  5y3 - 20x2y = 5y · (y2 - 4x2) = 5y [y2  - (2x)2] = 5y(y - 2x)(y + 2x),
          c)   9x2  - (x + 2)2 = [3x - (x + 2)] · [3x + (x + 2)] = (2x -2) · (4x + 2) = 4(x -1) · (2x +1).
Factoring quadratic trinomials
A quadratic trinomial  ax2 + bx + c  can be factorized as
ax2 + bx + c = a·[x2 + (b/a)·x + c/a] = a·(x - x1)(x - x2),  where x1 + x2 = b/a and  x1· x2 = c/a
11.    Factorize quadratic trinomials.
Solutions:   a)  x2 - 3x -10 = x2 + (-5 + 2)·x + (-5)·(+2) = x2 - 5x + 2x -10 =
                         = x · (x - 5) + 2 · (x - 5) = (x - 5) · (x + 2),
b)  2x2 - 7x + 3 = 2 · (x2 - 7/2x + 3/2) =  2(x2 - 1/2x - 3x + 3/2) =
                        = 2[x·(x - 1/2) - 3· (x - 1/2)] = 2· (x - 1/2)·(x - 3) = (2x - 1)· (x - 3),
c)  3x2 - x - 2 = 3(x2 - 1/3x - 2/3) =  3(x2 + 2/3x - x - 2/3) =
                        = 3[x·(x + 2/3) - (x + 2/3)] = 3·(x + 2/3)·(x - 1) = (3x + 2)·(x - 1).
12.    Given are leading coefficient a2 = -1 and the complex roots,  x1 = 1 + i  and  x2 = 1 - i, of a
second degree polynomial, find the polynomial using the above theorem.
Solution:   By plugging the given values into    a2x2 + a1x + a0 = a2(x - x1) (x - x2)
  a2x2 + a1x + a0 = -1[x - (1 + i)] · [x - (1 - i)] = -[(x - 1) - i] · [(x - 1) + i]
                           = -[(x - 1)2 - i2] = -(x2 - 2x + 1 + 1)  = - x2 + 2x - 2.
Factoring cubic or a third degree polynomial
  a3x3 + a2x2 + a1x + a0 = a3(x - x1) (x - x2) (x - x3) =   
  = a3[x3 - (x1 + x2 + x3)x2 + (x1x2 + x1x3 + x2x3)x - x1x2x3].
13.    The real root of the polynomial  - x3 - x2 + 4x - 6  is  x1 = - 3, factorize the polynomial.
Solution: We divide given polynomial by one of its known factors, 
a3x3 + a2x2 + a1x + a0 = a3(x - x1) (x - x2) (x - x3)
then we calculate another two roots of given cubic by solving obtained quadratic trinomial,
Finally we use the theorem to factorize given polynomial (see the previous example),
a3(x - x1)(x - x2)(x - x3) = -1(x + 3)[x - (1 + i)][x - (1 - i)]
                                         = - (x + 3)(x2 - 2x + 2).
Notice that given cubic has one real root and the pair of the conjugate complex roots.
Odd degree polynomials must have at least one real root.
Sum and difference of cubes
14.    Factorize given sum and difference of cubes.
Solutions:   a)  x3 + 8 = x3 + 23 = (x + 2) · (x2 - 2x + 22),
                        since  (x + 2)·(x2 - 2x + 4) = x3 - 2x2 + 4x + 2x2 - 4x + 8 = x3 + 8,
b) 8a3 -125 = (2a)3 - 53 = (2a - 5)· [(2a)2 + (2a)·5 + 52] = (2a - 5)(4a2 + 10a + 25),
  since  (2a - 5)(4a2 + 10a + 25) = 8a3 + 20a2 + 50a - 20a2 - 50a -125 = 8a3 -125.
Using a variety of methods including combinations of the above to factorize expressions
15.    Factorize given expressions.
Solutions:   a)  x2 - 2xy + y2 + 2y - 2x = (x - y)2 - 2(x - y) = (x - y)(x - y - 2),
b)  x2 - y2 + xz - yz = (x - y)(x + y) + z(x - y) = (x - y)(x + y + z),
c)  4x2  - 4xy  + y2  - z2 = (2x - y)2   - z2 = (2x - y - z)(2x - y + z),
d)  a3  - 7a + 6 = a3  - a - 6a + 6 = a(a2 -1) - 6(a -1) = (a -1)·[a(a + 1) - 6] = (a -1)(a2 + a - 6) =
                             = (a -1)(a2 + 3a - 2a - 6) = (a -1)[a(a + 3) - 2(a + 3)] = (a -1)(a + 3)(a - 2).
  Factoring and expanding algebraic expressions, rules for transforming algebraic expressions
Expanding algebraic expressions
The square of a binomial, a perfect square trinomial
  (a + b)2 = a2 + 2ab + b2,
  (a - b)2 =  a2 - 2ab + b2,
The square of a trinomial
  (a - b + c)2 = a2 + b2 + c2 - 2ab + 2ac - 2bc,
The cube of a binomial
  (a + b)3 = a3 + 3a2b + 3ab2 + b3,
  (a - b)3 = a3 - 3a2b + 3ab2 - b3,
Factoring algebraic expressions
Difference of two squares
  x2 - y2 = (x - y) · (x + y),
Sum and difference of cubes
  x3 - y3 = (x - y) · (x2 + xy + y2),
  x3 + y3 = (x + y) · (x2 - xy + y2).
The sum and/or difference of any two numbers raised to the same (positive integer) power
x4 - y4 = (x - y) · (x3 + x2y + xy2 + y3) = (x2 - y2) · (x2 + y2) 
x2n - y2n = (x - y) · (x2n-1 + x2n-2y + . . . + xy2n-2 + y2n-1) = (xn - yn) · (xn + yn) 
The binomial expansion algorithm - the binomial theorem
The binomial expansion of any positive integral power of a binomial, which represents a polynomial with n + 1 terms, 
or written in the form of the sum formula
 
 
is called the binomial theorem.
The binomial coefficients can also be obtained by using Pascal's triangle.
The triangular array of integers, with 1 at the apex, in which each number is the sum of the two numbers above it in the preceding row, as is shown in the initial segment in the diagram, is called Pascal's triangle.
So, for example the last row of the triangle contains the sequence of the coefficients of a binomial of the 5th power.
n 1
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
- 1 - - - - - 1
16.    Expand the binomial using the binomial theorem.
Solution:
 
 
 
 
 
Solved problems contents
Copyright © 2004 - 2012, Nabla Ltd.  All rights reserved.