THINKING FORMULAE

HOW TO IMPROVE YOUR THINKING

Thinking is the highest mental activity present in man. All human achievements and progress are simply the products of thought. The evolution of culture, art, literature, science and technology are all the results of thinking.

Thought and action are inseparable - they are actually the two sides of the same coin. All our deliberate action starts from our deliberate thinking. For a man to do something, he should first see it in his mind's eye -- he should imagine it, think about it first, before he can do it. All creations-- whether artistic, literal or scientific --first occur in the creator's mind before it is actually given life in the real world.

Misconceptions Regarding I.Q

·         The present education system develops and enhances thinking and so the more educated you are the better thinker you are.

·         Less Educated or Uneducated can never become good thinkers.

·         IQ and thinking ability are the same. The more IQ one has, the better thinking ability one has. On the contrary, those who have lower IQ have only low thinking abilities

·         Thinking ability, decision making ability and problem solving ability are inherent and there is very little we can do to develop these.

THINKING FORMULAE

some thinking formulae to help us practice thinking as a skilll. Most important of them are:

1. AGO

2. CAF

3. PMI

4. OPV

5. APC

AGO

AGO stands for Aims Goals Objectives. Since deliberate thinking is actually the manifestation of deliberate use of Will Power, it is important that the thinker should be well aware of the aims, goals and objectives. In other words, a sense of direction is required if one is to use his thinking effectively.

CAF

CAF stands for Consider All Factors. It reminds us that all factors or parameters of a problem should be considered to analyze it. By doing so, one can avoid the error of partialism in thinking.

PMI

PMI is the abbreviation for Plus Minus Interesting. When making decisions, this technique is very useful. First write down all the plus (i.e., positive) suggestions or aspects of a solution. Then write down all the minus (i.e., negative) aspects. Lastly, write down the interesting ideas or suggestions or aspects of the same. Now it is easy to arrive at the best suitable solution of the issue or problem at hand.

OPV

OPV means Other People's Views. In this technique, the thinker thinks from the perspectibve of the different people involved in or affected by the decision or solution. For example, a change in syllabus mainly affects the students, teachers, management and parents either directly or indirectly. So, before implementing a new syllabus, we should think from the perspective of all these people. This is especially important when the decision is enacted upon and through other people.

APC

APC is the short form for Alternatives Possibilities Choices. In this technique, the thinker generates as much alternatives, possibilities and choices for the solution of the problem. The best suitable one can then be selected by applying PMI or OPV.

HCF AND LCM OF POLYNOMIAL

HCF AND LCM OF POLYNOMIAL

HCF of Polynomials

The HCF of two or more given polynomials is the polynomial of highest degree and largest numerical coefficients which is a factor of all the given polynomials.

Method to find HCF of polynomial

The following method is suggested for finding the HCF of several polynomials

(a)    Write each polynomial as a product of prime factors.

(b)    The HCF is the product obtained by taking each factor to the lowest power to which it occurs in both the polynomials. For example,

The HCF of 23 32 (x - y)3(x + 2y)2; 2233 (x - y)2 (x + 2y)3 and 32(x - y)2 (x + 2y) is 32(x - y)2 (x + 2y).

Note-  Two or more polynomials are relatively prime if their HCF is 1.

LCM of Polynomials

The LCM of two or more given polynomials is the polynomial of lowest degree and smallest numerical coefficients for which each of the given polynomials will be its factor.

Method to find LCM of polynomial

The following procedure is suggested for determining the LCM of several polynomials.

(a)   Write each polynomial as a product of prime factors.

(b)   The LCM is the product obtained by taking each factor to the highest power to which it occurs. For example, The LCM of 2332 (x — y)3(x + 2y)2; 2233(x — y)2 (x + 2y)3; 32(x — y)2(x + 2y) is 2333 (X — y)3 (x + 2y)3.

Example1. Find the HCF and LCM of (I) 9X4y2 and 12X3y3

(ii) 6x — 6y and 4x2 — 4y2.

Sol. (i) 9x4y2 = 32 x x4 x y2 and

12X3y3 = 22x 3 x X3 x y3.

HCF = 3x3y2 and

LCM = 32 x 22 X x4 X y3 =36X4y3

So, HCF = 3x3y2, LCM = 36x4y3

(ii) 6x — 6y = 2 x 3 x (x — y) and

4x2 — 4y2 = 22(x2— y2) = 22 x (x + y) (x — y) HCF = 2(x — y) and

LCM = 22x 3 x (x — y)(x + y) = 12(x — y) (x + y). So, HCF = 2(x — y), LCM = 12(x + y) (x — y).

Special Products

The following are some of the products which occur frequently in mathematics and the student should become familiar with them as soon as possible.

Product of a monomial and a binomial a (c + d) = ac + ad.

Product of the sum and the difference of two terms (a + b)(a — b) = a2 — b2.

Square of a binomial                                          

(a + b)2 = a2 + 2ab + b2.

(a — b)2 = a2 — 2ab + b2

Product of two binomials

(x + a)(x + b) = x2 + (a + b)x + ab.

(ax + b)(cx + d) = acx2 + (ad + bc)x + bd. ,(a+b)(c+d)=ac+bc+ad+bd.

Cube of a binomial

(a + b)3 = a3 + 3a2b + 3ab2 + b3.

(a — b)3 = a3 — 3a2b + 3ab2 — b3.

Square of a trinomial

(a + b + c)2 = a2 + b2 + c2 + 2(ab + ac + bc).

Products that give answers of the form a n ± bn.

It may be verified by multiplication that (a — b)(a2 + ab + b2) = a3 — b3

(a — b)(a3 + a2b + ab2 + b3) = a4 — b4

(a — b)(a4 + a3b + a2b2 + ab3 + b4) = a5 — b5

(a — b) (a 5 + a 4 b + a 3  b 2 + a 2 b 3 + a b 4 + b 5 ) = a 6 — b 6  and so on These may be summarized as (a — b)(an-1 + an-2b + an-3 b2 + ... + ab2 + bn-1) = an — bn where n is any positive integer (1, 2, 3, 4, ...). Similarly, it may be verified that (a + b)(a2 - ab + b2) = a3 + b3

(a + b)(a4 — a3b + a2b2 — ab3 + b4) = a5 + b5

(a + b)(a6 — a5b + a4b2 — a3 b3 + a2 b4 — abs + b6) = a7 + b7 and so on.

These may be summarized as

(a + b)(an-1- an-2b + an-3 b2 — ... — abn-2 + bn-1) = an + bn  Where n is any positive odd integer (1, 3, 5, 7, ...).

Example2 . Simplify (3xy + 1)(2x2 — 3y).

Sol. (3xy)(2x2) + (3xy)(-3y) + (1)(2x2) + (1)(-3y) = 6x3y — 9xy2 + 2x2 — 3y.

Example3 .Simplify (x + y + z + 1)2.

Sol. [(x + y) + (z + 1)]2

= (x + y)2 + 2(x + y)(z + 1) + (z + 1)2

= x2 + 2xy + y2 + 2x2 + 2x + 2yz + 2y + z2 + 2z + 1.

Division Method

We use the division method to find HCF and LCM of monomials or binomials or trinomials or any algebraic expressions

 Problem 1: Find HCF and LCM of  (p+3)3, 2p3+54+18p(p+3), (p2+6p+9)

Solution:

Step 1: Find factors of all the expressions first.

1.The factors of (p+3)3 are (p+3),(p+3) and (p+3)

2. Let us simplify the 2nd term:

2p3+54+18p(p+3)

= 2(p3+27)+18p(p+3)

= 2*(p+3)( p2+9-3p)+18p(p+3),  (p3+27 is of the form a3+b3 with a=p and b=3we can apply the formula a3+b3 =(a+b) (a2 +b2 -ab)

=(p+3)*((2*(p2+9-3p))+18p)

= (p+3) *2*( p2+9-3p+9p)

=2(p+3)( p2+9+6p) ( (p2+9+6p) is of the form ( a2+ b2+2ab) with a=p and b=3, but ( a2+ b2+2ab)= (a+b)2

= 2(p+3)(p+3)2

 The factors of 2p3+54+18p(p+3) are 2, (p+3),(p+3),(p+3)

3. we have already seen above that (p2+6p+9) =(p+3)2

 The factors of (p2+6p+9) are p+3, p+3

Follow the division method to find HCF and LCM

The given terms can be rewritten as ( p+3)(p+3)(p+3),  2(p+3)(p+3)(p+3), (p+3)(p+3)

The first common factor is p+3 and let us start dividing terms by this term

(p+3) | ( p+3)(p+3)(p+3),  2(p+3)(p+3)(p+3), (p+3)(p+3)

(p+3) | (p+3)(p+3),            2(p+3)(p+3),          (p+3)

           (p+3),                     2(p+3)                            1

We stop further division as there are no more common factors for all the terms

Therefore HCF = (p+3)(p+3)= (p+3)2  and

(p+3) | ( p+3)(p+3)(p+3),  2(p+3)(p+3)(p+3), (p+3)(p+3)

(p+3) | (p+3)(p+3),            2(p+3)(p+3),          (p+3)

(p+3) | (p+3),                     2(p+3)                            1

              1,                          2,                                    1            

We stop further division as there are no more common factors for any 2 terms

Therefore LCM = (p+3)(p+3)(p+3)*1*2*1 = 2(p+3)3

Verification:

Let us cross check the solution by substituting p=2 in the above problem.

From the solution the HCF is (p+3)2 = (2+3)2 =25 and LCM= 2(p+3)3= 2(2+3)3= 2*125=250

Let us find the HCF and LCM of the given terms after converting them to numbers by substituting p=2.

Therefore the terms are  (2+3)3, (2*23+54+18*2(2+3)), (22+6*2+9)

= {125, 250,25}

By close observation we notice that HCF=25 and LCM=250

Since both the methods give same HCF and LCM our solution is correct.

 Problem 2: Find HCF and LCM of  10(x2-y2), 15(x2-2xy+y2) 20(x3- y3),5(-3x +3y)

Solution:

Step 1: Find factors of all the expressions first.

1. The first term has an expression of the form  (a2-b2) whose factors are (a+b) and (a-b) with a=x and b= y

 The factors of first term are 10, (x+y) and (x-y)

 10(x2-y2)=10(x+y)(x-y)

2. The second term has an expression of the form (a2-2ab+b2) whose factors are (a-b) and (a-b) with a=x and b= y

 The factors of second term are 15, (x-y) and (x-y)

 15(x2-2xy+y2)= 15(x-y) (x-y)

3. The third term has an expression of the form (x3-y3) whose factors are (x-y) and (x2 +y2 +xy) with a=x and b=y

 The factors of third term are  20, (x-y) and (x2 +y2 +xy)

4. The fourth term can be rewritten as 5*-3(x-y)

 The factors of fourth term are -15, (x-y)

5*-3(x-y) = 5*(-3)(x-y)=-15, (x-y) 

Step 2: Follow the division method to find HCF and LCM

The common factors are 5 and (x-y) so let us start dividing terms by these two together

5 (x-y) | 10(x+y) (x-y), 15(x-y) (x-y), 20(x-y)(x2 +y2 +xy), -15(x-y)

             2(x+y),             3(x-y),             4(x2 +y2 +xy),        -3

We stop further division as there are no more common factors for all the terms

Therefore HCF = 5(x-y)

To find LCM, we start division with 5(x-y)

5(x-y) | 10(x+y) (x-y), 15(x-y) (x-y), 20(x-y)(x2 +y2 +xy), -15(x-y)

        2|  2(x+y),             3(x-y),             4(x2 +y2 +xy),        -3 (We continue division as some terms have common factors)

        3|   (x+y),             3(x-y),              2(x2 +y2 +xy),        -3

   (x+y),                (x-y),              2(x2 +y2 +xy)         -1

We stop further division as there are no common factors among any 2 terms

Therefore LCM =5(x-y)* 2*3*(x+y)*(x-y)*2(x2 +y2 +xy)

=  60*(x-y)(x+y)*(x-y)(x2 +y2 +xy) (Note that (x-y)(x2 +y2 +xy) is of the form (a-b)( (a2 +b2 +ab)   with a=x and b= y)

= 60*(x2-y2)* (x3-y3)

Verification:

Since it is very difficult to cross verify easily, we will cross check the solution for at least  for one value of x and y by substituting x=3 and y=2 in the above problem

From the solution the HCF is 5(x-y) = 5*(3-2) = 5  and

LCM= 60*(x2- y2)* (x3-y3) = 60*(9-4)*)(27-8)=60*5*19=5700

Let us find the HCF and LCM of the given terms after converting them to numbers by substituting x=3 and y=2 in

10(x2-y2), 15(x2-2xy+y2) 20(x3- y3),5(-3x +3y)

Therefore the terms are 10(32-22), 15(32-2*3*2+22), 20(33- 23),5(-3*3 +3*2)

= {50, 15, 380, -15}

By observation we note that HCF=5

Let us use the division method to find LCM

5 | 50,15,380,-15

 2 | 10,3,76,-3

  3 | 5,3,38,-3

     |  5,1,38,-1

LCM = 5*2*3*5*38=5700

Since both the methods give same HCF and LCM our solution is not incorrect.

 Problem 3 : For what value of  a and b the polynomials

p(x) = (x2+3x+2) (x2+2x+a) and

q(x) = (x2+7x+12) (x2+7x+b)

have (x+1)(x+3) as their HCF

Solution:

(x2+3x+2) = (x+1)(x+2)

(x2+7x+12) = (x+4)(x+3)

 p(x) = (x+1)(x+2)(x2+2x+a)

q(x) = (x+4)(x+3) (x2+7x+b)

Since it is given that (x+1)(x+3) is HCF of  p(x),we conclude that

(x+3) is factor of (x2+2x+a)

This implies x=-3 satisfies the equation (x2+2x+a) =0

 (-3)2+2(-3)+a =0

I.e. 9-6+a =0

a =-3

Since it is given that (x+1)(x+3) is HCF of  q(x),we conclude that

(x+1) is factor of (x2+7x+b)

This implies x=-1 satisfies the equation (x2+7x+b) =0

 (-1)2+7(-1)+b =0

I.e. 1-7+b =0

b =6

Verfication :  By substituting value for a an b in p(x) and q(x) we get

p(x) = (x2+3x+2) (x2+2x-3) = (x+1) (x+2) (x+3) (x-1) { (x2+2x-3) = (x+3)(x-1)}

q(x) =(x2+7x+12) (x2+7x+6) = (x+4) (x+3) (x+1) (x+6) { (x2+7x+6)= (x+1)(x+6)}

By looking at factors of p(x) and q(x) we conclude that HCF of p(x) and q(x) is (x+1) (x+3)

Relationship between H.C.F. and L.C.M. of Two Polynomials

If f(x) and g(x) are two polynomials, then

           f(x). g(x) = {HCF of f(x) and g(x)} x {L.C.M of f(x) and g(x)}.               

Example: Find the L.C.M of the following pairs of polynomials with the help of their H.C.F.:

3 (x2 - 7x + 12) and 24(x2 - 9x + 20)

Solution:

Given, p(x) = 3 (x2 - 7x + 12)

= 3(x - 3)(x - 4)

and, q(x)  = 24(x2 - 9x + 20)

= 23 x 3 x (x - 4)(x - 5)

Clearly, H.C.F. of p(x) and q(x) is 3 (x - 4).

Therefore,

L.C.M. of p(x) and q(x)

= [p(x). q(x)] / H.C.F. of p(x) and q(x)

= { [3 (x - 3) (x - 4)1 x [23 x 3 x (x - 4) (x- 5)]}/ 3 (x - 4)

= (x - 3) x 23 x 3 x (x - 4) (x - 5)

= 24(x - 3)(x - 4)(x - 5)

 FREE HAND EXERCISE QUESTIONS

1. Find the HCF of the following:

(a) 2x4 - 2y4 and 3x3 + 6x2y - 3xy2 - 6y3

(b) 12(x3 + x2 + x + 1) and 18(x4 - 1)

(c) x3 + 2x2 - 3x and 2x3 + 5x2 - 3x

(d) 2(x4 - y4) and 3(x3 + 2x2y - xy2 - 2y3)

(e) 18(x3 - x2 + x - 1) and 12(x4 - 1)

(f) 45(x4 - x3 - x2) and 75(8x5 + x2)

(g) 36(3x4 + 5x3 - 2x2) and 54(27x4 - x)

(h) 42(2x3 - 5x2 - 3x) and 60(8x4 + x)

(i) 4(x4 - 1) and 6(x3 - x2 - x + 1)

2. Find the LCM of the following polynomials:

(a) 35(x4 - 27x) and 40(2x3 - 5x2 - 3x)

(b) 20(2x3 + 3x2 - 2x) and 45(x4 + 8x)

(c) 15(4x3 - 4x2 + x) and 35(2x2 - 7x + 3)

(d) 25(x2 + 7x + 12) and 15x(x2 - 16)

(e) x(8x3 + 27) and 2x2 (2x2 + 9x + 9)

3. Find the GCD and LCM of the polynomials P(x) and Q(x), where

P(x) = (x3 - 27) (x2 - 3x + 2) and

Q(x) =( x2 + 3x + 9) (x2 - 5x + 6)

4. The LCM and GCD of two polynomials, P(x) and Q(x) are 56(x4 + x) and 4(x2 - x + 1) respectively. If P(x) = 28(x3 + 1), find Q(x) .

5. For what value of k, the g.c.d. of x2 + x - (2k + 2) and 2x2 + kx - 12 is x + 4?

6. Find the value of K for which the g.c.d. of x2 - 2x - 24 and x2 - kx - 6 is x - 6.

7. (x2 - x - 6) is the GCD of the expression (x + 2) (2x2 + ax + 3) and (x - 3) (3x2 + bx + 8). Find the value of a and b.

8. (x + 1) ( x - 4) is the g.c.d. of the polynomials ( x - 4) (2x2 + x - a) and ( x + 1) (2x2 + bx - 12) find a and b .

9. (x - 3) is the g.c.d. of (x3 -2x2 + px + 6) and ( x2 - 5x + q) . Find (6p + 5q) .

10. Find the value of a and b so that the polynomials P(x) and Q(x) have (x - 1) (x + 4) as their HCF:

P(x) = (x2 - 3x + 2) (x2 + 7x + a)

Q(x) = (x2 + 5x + 4) (x2 - 5x + b)

11. Find the value of a and b so that the polynomial x3 + ax2 + bx + 15 is divisible by x2 + 2x - 15.

12. Find the value of p and q so that the polynomial f(x) = px3 + 2x2 - 19x + 9 is divisible by x2 + x - 6.

13. Determine the value of k such that x + 3 is a factor of the polynomial

f(x) = kx3 + x2 - 22x - 21

14. If x - 2 is a factor of x2 + ax - 6 = 0 and x2 - 9x + b = 0, find the value of a and b.Answers

ANSWERS

1. (a) x2 - y2                          (b) 6(x + 1) (x2 + 1)

    (c) x(x + 3)                         (d) x2 - y2

    (e) 6(x - 1) (x2 + 1)           (f) 15x2(2x + 1)

    (g) 18x(3x - 1)                   (h) 6x(2x + 1)

    (i) 2(x2 - 1)

2. (a) 280x(x - 3) (2x + 1) (x2 + 3x + 9)

    (b) 180x(x + 2) (2x - 1) (x2 - 2x + 4)

    (c) 105x((x - 3) (2x - 1)2

    (d) 75x(x2 - 16) (x + 3)

    (e) 2x2(8x2 + 27) (x + 3)

3. G. C. D. = (x3 - 27) (x - 2)

     LCM = (x - 1) (x - 2) (x - 3) (x2 + 3x + 9)

4. 8(x3 - x2 + x)

5. K = 5

6. K = 5

7. a = -7, b = 10

8. a = 1, b= -5

9. 0

10. a = 12, b = 4

11. a = 1, b = -17

12. p = 3, q = 6

13. K = 2              

14. a = 1, b = 14

Division algorithm for Polynomial

Division algorithm for Polynomial

Let us consider two numbers a and b such that a is divisible by b then a is called is dividend, b is called the divisor and the resultant that we get on dividing a with b is called the quotient and here the remainder is zero, since a is divisible by b. Hence by division rule we can write,

Dividend = divisor x quotient + remainder.

This holds good even for polynomials too. Let f(x), g(x), q(x) and r(x) are polynomials then the division algorithm for polynomials states that “If f(x) and g(x) are two polynomials such that degree of f(x) is greater that degree of g(x) where g(x) ≠ 0, then there exists unique polynomials q(x) and r(x) such that f(x) = g(x).q(x) + r(x) where r(x) = 0 or degree of r(x) less than degree of g(x)”.

RELATIONSHIP BETWEEN ZEROS AND COEFFICIENT OF A POLYNOMIAL

RELATIONSHIP  BETWEEN ZEROS AND COEFFICIENT OF A POLYNOMIAL

relationship  between zeros and coefficient of a polynomial in case of quadratic and cubic polynomial is stated as follows

(1) QUADRATIC POLNOMIAL

Let ax² +bx +c be the quadratic polynomial and α and β are its zeros ,then

Sum of zeros = α + β = -b/a = - (coefficient of x)/ (coefficient of x²)

Product of zeros = αβ = c/a = constant term / (coefficient of x²)

If we need to form an equation of degree two ,when sum and products of the roots is given ,then K[x²-( α + β )x + αβ ]=0 is the required equation ,where k is constant .

Procedure for finding zeros of a quadratic polynomial  

·         Find  the factors of the quadratic polynomial .

·         Equate each of the above factors (step 1) with zero.

·         Solve the above equation (step 2)  

·         The value of the variables obtained (step 3) are the required zeros .

(2) CUBIC POLYNOMIAL

Let axᶟ +bx² +cx +d  be the cubic polynomial and α , β and γ are its zeros ,then

 Sum of zeros = α + β + γ = -b/a = - (coefficient of x²)/ (coefficient of xᶟ)

Sum of Product of zeros taken two at a time  = αβ +βγ +γα = c/a =  (coefficient of x)/ (coefficient of xᶟ)

Product of zeros = αβγ = -d/a = - constant term / (coefficient of xᶟ)

When sum of zeros , Sum of Product of zeros taken two at a time , Product of zeros is given , then K[xᶟ-( α + β + γ )x² + (αβ +βγ +γα)x – αβγ ]=0 is the required equation ,where k is constant,

Procedure for finding zeros of a cubic polynomial

·         By  hit and trial method find one zeros of the polynomial using remainder theorem

·         Now if we know one zero , then we know one factor of the polynomial . divide the cubic polynomial by this factor to obtain quadratic polynomial

·         Now , solve this quadratic polynomial to obtain the other two zeros of the cubic polynomial .

·         These three zeros are the required zreos

ZEROS OF POLYNOMIAL

GEOMETRIC  MEANING OF ZEROS OF POLYNOMIAL

The zero of the polynomial is defined as any real value of x, for which the value of the polynomial becomes zero.

A real number k  is a zero of a polynomial p(x), if .p(k) =0

Geometrical Meaning of the Zeroes of a Polynomial: The zero of the polynomial is the x-coordinate of the point, where the graph intersects the x-axis. If a polynomial p(x)  intersects the x-axis at (k,0) , then k  is the zero of the polynomial.

The graph of a linear polynomial intersects the x-axis at a maximum of one point. Therefore, a linear polynomial has a maximum of one zero.

The graph of a quadratic polynomial intersects the x-axis at a maximum of two points. Therefore, a quadratic polynomial can have a maximum of two zeroes. In case of a quadratic polynomial, the shape of the graph is a parabola. The shape of the parabola of a quadratic polynomial ax² +bx +c , a≠ 0  depends on ‘a’ .

If  a> 0 , then the parabola opens upwards.

If a < 0 , then the parabola opens downwards.

The graph of a cubic polynomial intersects the x-axis at maximum of three points. A cubic polynomial has a maximum of three zeroes. In general, an nth-degree polynomial intersects the x-axis at a maximum of n points. Therefore, an nth-degree polynomial has a maximum of n zeroes.

GRAPHING POLYNOMIALS

Graphing Higher Degree Polynomials

As the degree of a polynomial increases, it becomes increasingly hard to sketch it accurately and analyze it completely. There are a few things we can do, though.

Using the Leading Coefficient Test, it is possible to predict the end behavior of a polynomial function of any degree. Every polynomial function either approaches infinity or negative infinity as x increases and decreases without bound. Which way the function goes as x increases and decreases without bound is called its end behaviour

If the degree of the polynomial function is even, the function behaves the same way at both ends (as x increases, and as x decreases). If the leading coefficient is positive, the function increases as x increases and decreases. If the leading coefficient is negative, the function decreases as x increases and decreases.

If the degree of the polynomial function is odd, the function behaves differently at each end (as x increases, and as x decreases). If the leading coefficient is positive, the function increases as x increases, and decreases as x decreases. If the leading coefficient is negative, the function decreases as x increases and increases as x decreases.

If the leading coefficient test gets confusing, just think of the graphs of y = x 2 and y = - x 2 , as well as y = x 3 and y = - x 3 . The behavior of these graphs, which hopefully by now you can picture in your head, can be used as a guide for the behavior of all higher polynomial functions.

Besides predicting the end behavior of a function, it is possible to sketch a function, provided that you know its roots. By evaluating the function at a test point between roots, you can find out whether the function is positive or negative for that interval. Doing this for every interval between roots will result in a rough, but in many ways accurate, sketch of a function.



POLYNOMIAL

Polynomials

 Definition:

A polynomial is a monomial or the sum or difference of monomials. Each monomial is called a term of the poynomial

Terms

Ascending Order  -  An ordering of terms of a polynomial in which the degree of each term is at least as large as the degree of the preceding term.

Binomial  -  A polynomial with two terms.

Common Factor  -  A factor of all the terms in an expression.

Degree  -  of a monomial. The sum of the exponents of its variables. of a polynomial. The largest of the degrees of its individual terms.

Descending Order  -  An ordering of terms of a polynomial in which the degree of each term is no larger than the degree of the preceding term.

Difference of Squares  -  The binomial which is the product of the sum and difference of the same two terms, i.e. one of the form a 2 - b 2 .

Factor  -  A factor is integer, variable, or polynomial that can be multiplied by a constant, an integer, or a polynomial to produce the given expression. To factor means to separate an expression into simpler factors. For instance,

a 2 -2ab + b 2      =             (a - b)(a - b)       

ax 2 + abx + ac   =             a(x 2 + bx + c)   

Leading Coefficient  -  The coefficient of the first term of a polynomial written in descending order.

Monomial  -  An expression with a single term; a real number, a variable, or the product of real numbers and variables

Perfect Square Trinomial  -  The square of a binomial; has the form a 2 +2ab + b 2 .

Polynomial  -  A monomial or the sum or difference of several monomials.

Trinomial  -  A polynomial with three terms.

Classification of Polynomials

Polynomials are classified according to two attributes -- number of terms and degree.

Classification of Polynomials by Number of Terms

A monomial is an expression with a single term. It is a real number, a variable, or the product of real numbers and variables. For example, 4 , 3x 2 , and 15xy 3 are all monomials, but 4x 2 + x , (3 + y)2 , and 12 - z are not monomials.

A polynomial is a monomial or the sum or difference of monomials. 4x 3 +3y + 3x 2 + z , -12zy , and 15 - x 2 are all polynomials.

Polynomials are classified according to their number of terms. 4x 3 +3y + 3x 2 has three terms, -12zy has 1 term, and 15 - x 2 has two terms. As already mentioned, a polynomial with 1 term is a monomial. A polynomial with two terms is a binomial, and a polynomial with three terms is a trinomial.

Classification of Polynomials by Degree

The degree of a monomial is the sum of the exponents of its variables. For example, 12x 3 has degree 3, x 2 y 5 has degree 2 + 5 = 7 , and 11xy has degree 1 + 1 = 2 .

A polynomial can be arranged in ascending order, in which the degree of each term is at least as large as the degree of the preceding term, or in descending order, in which the degree of each term is no larger than the degree of the preceding term. The polynomial 3 + 12x - xy + 7x 2 y + y 5 -12x 3 y 3 is written in ascending order, while the same polynomial expressed as -12x 3 y 3 + y 5 +7x 2 y - xy + 12x + 3 is written in descending order.

Mathematicians generally write polynomials in descending order. The coefficient of the first term of a polynomial written in descending order is known as the leading coefficient.

The degree of a polynomial is the largest of the degrees of its monomial terms.

Addition and Subtraction of Polynomials

Addition of Polynomials

To add two polynomials, combine their like terms.

Example 1: (4x 5 + x 4 -12x 3 + x - 6) + (3x 4 +8x 3 +6x 2 - x) = ?

= 4x 5 + (1 + 3)x 4 + (- 12 + 8)x 3 +6x 2 + (- 1 + 1)x - 6

= 4x 5 +4x 4 -4x 3 +6x 2 + 0x - 6

= 4x 5 +4x 4 -4x 3 +6x 2 - 6

Example 2: (- 3x 4 -2x 2 + x + 17) + (14x 6 +12x 5 -2x 4 +7x 3 +2x 2 - 4x - 12) = ?

= 14x 6 +12x 5 + (- 3 - 2)x 4 +7x 3 + (- 2 + 2)x 2 + (1 - 4)x + (17 - 12)

= 14x 6 +12x 5 -5x 4 +7x 3 +0x 2 - 3x + 5

= 14x 6 +12x 5 -5x 4 +7x 3 - 3x + 5

Subtraction of Polynomials

To subtract one polynomial from another, change the subtraction sign to an addition sign and change the signs of all the terms in the polynomial being subtracted (don't forget to change the sign of the first term). Then add the resulting polynomial by combining like terms.

Example 3: (3x 4 +4x 3 -5x 2 +18) - (7x 5 +3x 4 -12x 3 -3x 2 + 7x - 9) = ?

= (3x 4 +4x 3 -5x 2 +18) + (- 7x 5 -3x 4 +12x 3 +3x 2 - 7x + 9)

= - 7x 5 + (3 - 3)x 4 + (4 + 12)x 3 + (- 5 + 3)x 2 - 7x + (18 + 9)

= - 7x 5 +16x 3 -2x 2 - 7x + 27

Example 4: (- 5x 6 -3x 4 +18x 3 -7x 2 +9x + 11) - (- x 4 + x 3 -12x 2 + 10x + 5) = ?

= (- 5x 6 -3x 4 +18x 3 -7x 2 +9x + 11) + (x 4 - x 3 +12x 2 - 10x - 5)

= - 5x 6 + (- 3 + 1)x 4 + (18 - 1)x 3 + (- 7 + 12)x 2 + (9 - 10)x + (11 - 5)

= - 5x 6 -2x 4 +17x 3 +5x 2 - x + 6

Multiplication of Polynomials

Multiplication of a Polynomial by a Monomial

To multiply a polynomial by a monomial, use the distributive property: multiply each term of the polynomial by the monomial. This involves multiplying coefficients and adding exponents of the appropriate variables.

Example 1: 3y 2(12y 3 -6y 2 + 5y - 1) = ?

= 3y 2(12y 3) + (3y 2)(- 6y 2) + (3y 2)(5y) + (3y 2)(- 1)

= (3)(12)y 2+3 + (3)(- 6)y 2+2 + (3)(5)y 2+1 + (3)(- 1)y 2

= 36y 5 -18y 4 +15y 3 -3y 2

Example 2: -4x 3 y(- 2y 2 + xy - x + 9) = ?

= - 4x 3 y(- 2y 2) + (- 4x 3 y)(xy) + (- 4x 3 y)(- x) + (- 4x 3 y)(9)

= (- 4)(- 2)x 3 y 1+2 + (- 4)x 3+1 y 1+1 + (- 4)(- 1)x 3+1 y + (- 4)(9)x 3 y

= 8x 3 y 3 -4x 4 y 2 +4x 4 y - 36x 3 y

Multiplication of Binomials

To multiply a binomial by a binomial-- (a + b)(c + d ) , where a , b , c , and d are terms--use the distributive property twice. First, treat the second binomial as a single term and distribute over the first binomial:

(a + b)(c + d )= a(c + d )+ b(c + d )                 

Next, use the distributive property over the second binomial:

a(c + d )+ b(c + d )= ac + ad + bc + bd          

At this point, there should be 4 terms in the answer -- every combination of a term of the first binomial and a term of the second binomial. Simplify the answer by combining like terms.

We can use the word FOIL to remember how to multiply two binomials (a + b)(c + d ) :

Multiply their First terms. (ac)

Multiply their Outside terms. (ad )

Multiply their Iinside terms. (bc)

Multiply their Last terms. (bd )

Finally, add the results together: ac + ad + bc + bd . Combine like terms.

Remember to include negative signs as part of their respective terms in the multiplication.

Example 1. (xy + 6)(x + 2y) = ?

= (xy)(x) + (xy)(2y) + (6)(x) + (6)(2y)

= x 2 y + 2xy 2 + 6x + 12y

Example 2. (3x 2 +7)(4 - x 2) = ?

= (3x 2)(4) + (3x 2)(- x 2) + (7)(4) + (7)(- x 2)

= 12x 2 -3x 4 +28 - 7x 2

= - 3x 4 + (12 - 7)x 2 + 28

= - 3x 4 +5x 2 + 28

Example 3: (y - x)(- 4y - 3x) = ?

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

= - 4y 2 -3xy + 4xy + 3x 2

= 3x 2 + (- 3 + 4)xy - 4y 2

= 3x 2 + xy - 4y 2

Multiplication of Polynomials

The strategy for multiplying two polynomials in general is similar to multiplying two binomials. First, treat the second polynomial as a single term, and distribute over the first term:

(a + b + c)(d + e + f )= a(d + e + f )+ b(d + e + f )+ c(d + e + f )           

Next, distribute over the second polynomial:

a(d + e + f )+ b(d + e + f )+ c(d + e + f )= ad + ae + af + bd + be + bf + cd + ce + cf                    

At this point, the number of terms in the answer should be the number in the first polynomial times the number in the second polynomial--every combination of a term of the first polynomial and a term of the second polynomial. Since there are 3 terms in each polynomial in this example there should be 3(3) = 9 terms in our answer so far. If the first polynomial had 4 terms and the second had 5 , there would be 4(5) = 20 terms in the answer so far.

Finally, since the the terms in such a product of polynomials are often highly redundant (many have the same variables and exponents), it is important to combine like terms.

Example 1: (x 2 -2)(3x 2 - 3x + 7) = ?

= x 2(3x 2 -3x + 7) - 2(3x 2 - 3x + 7)

= x 2(3x 2) + x 2(- 3x) + x 2(7) - 2(3x 2) - 2(- 3x) - 2(7) ( 6 terms)

= 3x 4 -3x 3 +7x 2 -6x 2 + 6x - 14

= 3x 4 -3x 3 + (7 - 6)x 2 + 6x - 14

= 3x 4 -3x 3 + x 2 + 6x - 14

Example 2: (x 2 + x + 3)(2x 2 - 3x + 1) = ?

= x 2(2x 2 -3x + 1) + x(2x 2 -3x + 1) + 3(2x 2 - 3x + 1)

= x 2(2x 2) + x 2(- 3x) + x 2(1) + x(2x 2) + x(- 3x) + x(1) + 3(2x 2) + 3(- 3x) + 3(1) ( 9 terms)

= 2x 4 -3x 3 + x 2 +2x 3 -3x 2 + x + 6x 2 - 9x + 3

= 2x 4 + (- 3 + 2)x 3 + (1 - 3 + 6)x 2 + (1 - 9)x + 3

= 2x 4 - x 3 +4x 2 - 8x + 3

Note: To check your answer, pick a value for the variable and evaluate both the original expression and your answer--they should be the same.

Multiplication of Binomials -- Special Cases

Square of a Binomial

To square a binomial, multiply the binomial by itself:

(a + b)2 = (a + b)(a + b)

(a + b)2 =             (a + b)(a + b)     

                =             a 2 + ab + ba + b 2             

                =             a 2 + ab + ab + b 2             

                =             a 2 +2ab + b 2      

The square of a binomial is always the sum of:

The first term squared,

2 times the product of the first and second terms, and

the second term squared.

When a binomial is squared, the resulting trinomial is called a perfect square trinomial.

Examples:

(x + 5)2 = x 2 +2(x)(5) + 52 = x 2 + 10x + 25

(100 - 1)2 = 1002 +2(100)(- 1) + (- 1)2 = 10000 - 200 + 1 = 9801

(2x - 3y)2 = (2x)2 +2(2x)(- 3y) + (- 3y)2 = 4x 2 -12xy + 9y 2

Product of the Sum and Difference of Two Terms

When we multiply two polynomials that are the sum and difference of the same 2 terms -- (x + 5) and (x - 5) for example -- we get an interesting result:

(a + b)(a - b)   = a(a) + a(- b) + ba + b(- b)              

                =             a 2 - ab + ab - b 2               

                =             a 2 - b 2                  

The product of the sum and difference of the same two terms is always the difference of two squares; it is the first term squared minus the second term squared. Thus, this resulting binomial is called a difference of squares.

Examples:

(7 - 2)(7 + 2) = 72 -22 = 49 - 4 = 45

(x + 9)(x - 9) = x 2 -92 = x 2 - 81

(2x - y)(2x + y) = (2x)2 - y 2 = 4x 2 - y 2

(3x 2 -2)(3x 2 +2) = (3x 2)2 -22 = 9x 4 - 4

(- y + 5x)(- y - 5x) = (- y)2 - (5x)2 = y 2 -15x 2

Removing Common Factors

Factors

A factor is a number that evenly divides the given number. A factor need not be a constant. In fact, any integer, variable, or polynomial that can be multiplied by an integer, a variable, or a polynomial to produce the given expression is a factor of the given expression.

Removing Common Factors

We've seen how to distribute a quantity over a polynomial and write the result as a polynomial. We can actually reverse this process--we can "remove" a common factor from a polynomial and write the result as a quantity times a polynomial. For example, 12 + 2x can be written as 2(6 + x) .

The first step to removing a common factor is finding a common factor. A common factor is a factor of all the terms in an expression (i.e., a factor that they all have in common). A common factor can be an integer, a variable, or a combination of integers and variables.

To remove a common factor and rewrite a polynomial as the product of a monomial and another polynomial:

Find the greatest common factor which is a whole number (no variables).

Divide all terms of the polynomial by that factor, and put the result in parentheses. Write the factor outside the parentheses.

Find the greatest common factor which is a variable or a product of several variables. That is, find the variables contained in every term, and write them with their lowest exponent.

Divide each term of the expression in parentheses by the greatest common variable factor, and write the variable factor outside the parentheses.

Check--distributing the monomial over the new polynomial should yield the original polynomial.

Example 1: Factor 4x 2 +16x 3 + 8x .

The greatest common whole number factor is 4 .

4x 2 +16x 3 +8x = 4(x 2 +4x 3 + 2x)

The greatest common variable factor is x ( x is contained in all the terms, and its lowest exponent is 1 ).

4(x 2 +4x 3 +2x) = 4x(x + 4x 2 + 2)

Check: 4x(x + 4x 2 +2) = 4x 2 +16x 3 + 8x

Thus, 4x 2 +16x 3 +8x = 4x(x + 4x 2 + 2) .

Example 2: Factor 12x 3 y + 3x 4 y 2 -6x 2 y 2 z .

The greatest common whole number factor is 3 .

12x 3 y + 3x 4 y 2 -6x 2 y 2 z = 3(4x 3 y + x 4 y 2 -2x 2 y 2 z)

The greatest common variable factor is x 2 y ( x is contained in all the terms, and its lowest exponent is 2 ; y is contained in all the terms, and its lowest exponent is 1; z is not contained in all the terms).

3(4x 3 y + x 4 y 2 -2x 2 y 2 z) = 3x 2 y(4x + x 2 y - 2yz)

Check: 3x 2 y(4x + x 2 y - 2yz) = 12x 3 y + 3x 4 y 2 -6x 2 y 2 z

Thus, 12x 3 y + 3x 4 y 2 -6x 2 y 2 z = 3x 2 y(4x + x 2 y - 2yz) .

Factoring Trinomials

Factoring Trinomials of the Form x 2 + bx + c and x 2 - bx + c

Just as the product of two binomials can often be rewritten as a trinomial, trinomials of the form ax 2 + bx + c can often be rewritten as the product of two binomials. For example, x 2 + 3x + 2 = (x + 1)(x + 2) .

We now know that the product of two binomials of the form (x + d ) and (x + e) is given by:

(x + d )(x + e) = x 2 + xe + dx + de = x 2 + (d + e)x + de        

Thus, in order to rewrite a binomial x 2 + bx + c as the product of two binomials ( b positive or negative), we must find numbers d and e such that d + e = b and de = c . Since c is positive, d and e must have the same sign.

Here are the steps to factoring a trinomial of the form x 2 + bx + c , with c > 0 . We assume that the coefficients are integers, and that we want to factor into binomials with integer coefficients.

Write out all the pairs of numbers which can be multiplied to produce c .

Add each pair of numbers to find a pair that produce b when added. Call the numbers in this pair d and e .

If b > 0 , then the factored form of the trinomial is (x + d )(x + e) . If b < 0 , then the factored form of the trinomial is (x - d )(x - e) .

Check: The binomials, when multiplied, should equal the original trinomial.

Note: Some trinomials cannot be factored. If none of the pairs total b , then the trinomial cannot be factored.

Example 1: Factor x 2 + 5x + 6 .

Pairs of numbers which make 6 when multiplied: (1, 6) and (2, 3) .

1 + 6≠5 . 2 + 3 = 5. Thus, d = 2 and e = 3 (or vice versa).

(x + 2)(x + 3)

Check: (x + 2)(x + 3) = x 2 +3x + 2x + 6 = x 2 + 5x + 6

Thus, x 2 + 5x + 6 = (x + 2)(x + 3) .

Example 2: Factor x 2 - 7x + 12 .

Pairs of numbers which make 12 when multiplied: (1, 12) , (2, 6) , and (3, 4) .

1 + 12≠7 . 2 + 6≠7 . 3 + 4 = 7. Thus, d = 3 and e = 4 .

(x - 3)(x - 4)

Check: (x - 3)(x - 4) = x 2 -4x - 3x + 12 = x 2 - 7x + 12

Thus, x 2 - 7x + 12 = (x - 3)(x - 4) .

Example 3: Factor 2x 3 +4x 2 + 2x .

First, remove common factors: 2x 3 +4x 2 +2x = 2x(x 2 + 2x + 1)

Pairs of numbers which make 1 when multiplied: (1, 1) .

1 + 1 = 2. Thus, d = 1 and e = 1 .

2x(x + 1)(x + 1) (don't forget the common factor!)

Check: 2x(x + 1)(x + 1) = 2x(x 2 +2x + 1) = 2x 3 +4x 2 + 2x

Thus, 2x 3 +4x 2 +2x = 2x(x + 1)(x + 1) = 2x(x + 1)2 .

x 2 + 2x + 1 is a perfect square trinomial.

Factoring Trinomials of the Form x 2 + bx - c and x 2 - bx - c

In the equation (x + d )(x + e) = x 2 + (d + e)x + de = x 2 + bx + c (mentioned in Heading ), c is negative if and only if de is negative; that is, if and only if d is negative or e is negative. Thus, the equation becomes (x + d )(x - e) = x 2 + (d - e)x - de = x 2 + bx - c , which c > 0 .

In order to rewrite the trinomial x 2 + bx - c as the product of two binomials, we must find numbers d and e such that d - e = b and | de| = c . If b is positive, then d - e > 0 , so d > e . If b is negative, then d - e < 0 , so d < e .

Here are the steps to factoring a trinomial of the form x 2 + bx - c , with c > 0 :

Write out all the pairs of numbers which can be multiplied to produce c .

Subtract each pair of numbers to find a pair that produce b when one is subtracted from the other. Call the numbers in this pair d and e . If b > 0 , let d be the larger number, and if b < 0 , let d be the smaller number.

The factored form of the trinomial is (x + d )(x - e) .

Check: The binomials, when multiplied, should equal the original trinomial. If the middle term has the wrong sign, you most likely switched d and e . Switch the "+" and "-" sign in your binomials and check again.

Note: Again, not every trinomial can be factored.

Example 1: Factor x 2 + 6x - 16 .

Pairs of numbers which make 16 when multiplied: (1, 16) , (2, 8) , and (4, 4) .

16 - 1≠6 . 8 - 2 = 6. Since b = 6 > 0 , d = 8 and e = 2 .

(x + 8)(x - 2)

Check: (x + 8)(x - 2) = x 2 -2x + 8x - 16 = x 2 + 6x - 16

Thus, x 2 + 6x - 16 = (x + 8)(x - 2) .

Example 2: Factor x 2 - x - 20 .

Pairs of numbers which make 20 when multiplied: (1, 20) , (2, 10) , and (4, 5) .

20 - 1≠1 . 10 - 2≠1 . 5 - 4 = 1. Since b = - 1 < 0 , d = 4 and e = 5 .

(x + 4)(x - 5)

Check: (x + 4)(x - 5) = x 2 -5x + 4x - 20 = x 2 - x - 20

Thus, x 2 - x - 20 = (x + 4)(x - 5) .

Example 3: Factor x 2 - 16 (Here, x has an implied coefficient of b = 0 ).

Pairs of numbers which make 16 when multiplied: (1, 16) , (2, 8) , and (4, 4) .

16 - 1≠ 0 . 8 - 2≠ 0 . 4 - 4 = 0. Thus, d = 4 and e = 4 .

(x + 4)(x - 4)

Check: (x + 4)(x - 4) = x 2 -4x + 4x - 16 = x 2 - 16

Thus, x 2 - 16 = (x + 4)(x - 4) .