tppt on ploynimial .by Tushar thapliyal

1. Power Point Presentation
on
POLYNOMIALS
By : Tushar Thapliyal
Class : IX – , ROLL NO . .

2. POLYNOMIAL
 Polynomial comes from poly- (meaning "many") and -
nomial (in this case meaning "term") ... so it says "many terms“.
 A polynomial can have:
 constants (like 3, -20, or ½)
 variables (like x and y)
 exponents (like the 2 in y2
), but only 0, 1, 2, 3, ... etc are allowed
 that can be combined using addition, subtraction,
multiplication and division ...
... except ...
 ... not division by a variable (so something like 2/x is right out).
 So: A polynomial can have constants, variables and
exponents, but never division by a variable.

3. Polynomial or Not?
POLYNIMIAL EXAMPLES
example of a polynomial, this one has 3 terms
These are polynomials:
•3x
•x – 2
•-6y2
- (7
/9
)x
•3xyz + 3xy2
z - 0.1xz - 200y + 0.5
•512v5
+ 99w5
•5
(Yes, even "5" is a polynomial, one term is allowed, and it can even be just a constant!)

4. And these are not polynomials
 3xy-2
is not,because the exponent is "-2" (exponents can only be 0,1,2,...).
 2/(x+2) is not, because dividing by a variable is not allowed.
 1/x is not either
 √x is not, because the exponent is "½" (see fractional exponents)
But these are allowed:
 x/2 is allowed, because you can divide by a constant also 3x/8 for
the same reason.
 √2 is allowed, because it is a constant (= 1.4142...etc).
Monomial, Binomial, Trinomial
There are special names for polynomials with 1, 2 or 3 terms:
Can Have Lots and Lots of Terms
 Polynomials can have as many terms as needed, but not an infinite number
of terms.
Variables : Polynomials can have no variable at all
 Example: 21 is a polynomial. It has just one term, which is a constant.
 one variable
Example: x4
-2x2
+x has three terms, but only one variable (x)
 Or two or more variables
 Example: xy4
-5x2
z has two terms, and three variables (x, y and z)

5. See how nice and
smooth the curve is?
The Degree is 3 (the largest
exponent of x)
What is Special About Polynomials?
Because of the strict definition, polynomials are easy to work with.
For example we know that:
• If you add polynomials you get a polynomial
• If you multiply polynomials you get a polynomial
 So you can do lots of additions and multiplications, and still have a polynomial as the
result.
 Also, polynomials of one variable are easy to graph, as they have smooth and
continuous lines.
Example: x4
-2x2
+x
You can also divide polynomials (but the result may not be a polynomial).
Degree :
The degree of a polynomial with only one variable is the largest exponent of that variable.
Example:
Standard Form :
The Standard Form for writing a polynomial is to put the terms with the highest degree first.
Example: Put this in Standard Form: 3x2
- 7 + 4x3
+ x6
The highest degree is 6, so that goes first, then 3, 2 and then the constant last:
x6
+ 4x3
+ 3x2
- 7
You don't have to use Standard Form, but it helps.

6. • Degreeof polynomial :- thehighest power of the
variablein apolynomial istermed asthedegreeof
polynomial.
• Constant polynomial :- A polynomial of degreezero is
called constant polynomial.
• Linear polynomial :- A polynomial of degreeone.
• E.g. :-9x + 1
• Quadratic polynomial :- A polynomial of degree two.
E.g. :-3/2y² -3y + 3
• Cubic polynomial :- A polynomial of degreethree.
• E.g. :-12x³ -4x² + 5x +1
• Bi – quadratic polynomial :- A polynomial of degree
four.
• E.g. :- 10x – 7x ³+ 8x² -12x + 20

7. • . Standard Form
• The Standard Form for writing apolynomial isto put
thetermswith thehighest degreefirst.
• Example: Put this in Standard Form: 3x2
- 7 +
4x3
+ x6
Thehighest degreeis6, so that goesfirst, then 3, 2
and then theconstant last:
x6
+ 4x3
+ 3x2
- 7

8. ZERO OF POLYNOMIAL
 It is a solution to the polynomial equation, P(x) = 0. It is that value of
x that makes the polynomialequal to 0. In other words, the number r
is a root of a polynomial P(x).
 We say that is a root or zero of a polynomial, P(x) , if P(r)=0 . In
other words, x=r is a root or zero of a polynomial if it is a solution to
the equation P(x)=0 .
 Let’s first find the zeroes for . To do this we simply
solve the following equation.
 So, this second degree polynomial has two zeroes or roots.
 Now, let’s find the zeroes for . That will mean solving,
 So, this second degree polynomial has a single zero or root. Also, recall
that when we first looked at these we called a root like this a double
root.
Ex:
 We’ve also got a product of three terms in this polynomial. However, since the first is now
an x this will introduce a third zero. The zeroes for this polynomial are,
because each of these will make one of the terms, and hence the whole polynomial, zero.

9. Multiplying Two Polynomials
Examples:
( ) ( )2
5 10 3x x x+ + − =
( )( )2
4 5 3 4x x x+ + − =
3
x 2
10x+ 3x− 2
5x+ 50x+ 15−
3
x 2
15x+ 47x+ 15−
3
12x 2
16x− 2
3x+ 4x− 15x+ 20− =
3
12x 2
13x− 11x+ 20−
Polynomials and Polynomial Functions
Multiplication

10. Dividing Polynomials
 In this section we’re going to take a brief look at dividing polynomials. This is something that we’ll be doing off and on
throughout the rest of this chapter and so we’ll need to be able to do this.
 Let’s do a quick example to remind us how long division of polynomials works.
 Example 1 Divide by .
 Solution
 Let’s first get the problem set up
 Recall that we need to have the terms written down with the exponents in decreasing order and to make sure we don’t make any mistakes we add in
any missing terms with a zero coefficient.

 Now we ask ourselves what we need to multiply to get the first term in first polynomial. In this case that is . So multiply by
and subtract the results from the first polynomial.
 The new polynomial is called the remainder. We continue the process until the degree of the remainder is less than the degree of the divisor, which
is in this case . So, we need to continue until the degree of the remainder is less than 1.
 Recall that the degree of a polynomial is the highest exponent in the polynomial. Also, recall that a constant is thought of as a polynomial of degree
zero. Therefore, we’ll need to continue until we get a constant in this case.
 Here is the rest of the work for this example.
 Okay, now that we’ve gotten this done, let’s remember how we write the actual answer down. The answer is,

11. Polynomials in one variable
 A polynomial in one variable is a function in which the variable is only to
whole number powers, and the variable does not appear in
denominators, in exponents, under radicals, or in between absolute
value signs or greatest integer signs.
 Examples of polynomials in one variable:
 −3x4
+ x3
− x √ 4 + 8 .
 P(x) = p0 + p1x + ...
+ pnxn.
 1 − 3/ 5 t7
.
 (x2
+ x + 1)(3x − 8).
 3 .
Examples of expressions that are not polynomials:
 3x2
− 3√x
 2 x
 x + 1/ 3x4
− 1

12. • Let p(x) beany polynomial of degree
greater than or equal to oneand let a
beany real number. If p(x) isdivided
by linear polynomial x-athen the
reminder isp(a).
• Proof :- Let p(x) beany polynomial of
degreegreater than or equal to 1. suppose
that when p(x) isdivided by x-a, the
quotient isq(x) and thereminder isr(x), i.g;
p(x) = (x-a) q(x) +r(x)
Remindertheorem

13. Sincethedegreeof x-ais1 and thedegreeof r(x) isless
than thedegreeof x-a,thedegreeof r(x) = 0.
Thismeansthat r(x) isaconstant .say r.
So , for every valueof x, r(x) = r.
Therefore, p(x) = (x-a) q(x) + r
In particular, if x = a, thisequation givesus
p(a) =(a-a) q(a) + r
Which provesthetheorem.

14. Example:
 2x2
-5x-1 divided by x-3
 f(x) is 2x2
-5x-1
 g(x) is x-3
 After dividing we get the answer 2x+1, but there is a remainder
of 2
 q(x) is 2x+1
 r(x) is 2
 In the style f(x) = g(x)·q(x) + r(x) we can write:
 2x2
-5x-1 = (x-3)(2x+1) + 2

15. Let p(x) beapolynomial of degree
n > 1 and let abeany real number.
If p(a) = 0 then (x-a) isafactor of
p(x).
PROOF:-By thereminder theorem
,
p(x) = (x-a) q(x) + p(a).
FactorTheorem

16. 1. If p(a) = 0,then p(x) = (x-a) q(x), which
showsthat x-aisafactor of p(x).
2. Sincex-aisafactor of p(x),
p(x) = (x-a) g(x) for samepolynomial
g(x).
In thiscase, p(a) = (a-a) g(a) =0

17. factorization of polynomialsFactor 12y2
– 5y.
In this case, no number is a common factor between the two terms (specifically,
the 12 and the5 share no common numerical factor), but I can still divide out a
common variablefactor of "y" from each of the two terms.
12y2
– 5y = y( )
In the first term, I have the "12" and the other "y" factor left over:
12y2
– 5y = y(12y )
(This is because 12y2
means 12×y×y, so taking the 12 and one of the y's out front leaves
the second y behind.) In the second term, I have the "5" left over:
12y2
– 5y = y(12y – 5)
Factor x2
+ 4x – x – 4.
This polynomial has four terms with no factor common to all four, so I'll try to factor "in
pairs":
x2
+ 4x – x – 4
= x(x + 4) – 1(x + 4)
= (x + 4)(x – 1)
In the second line above, I factored a "1" out. Why? If "nothing" factors out, a "1"
factors out
Factor x2
– 4x + 6x – 24.
I'll try to factor "in pairs": Copyright © Elizabeth Stapel 2002-2011 All Rights
Reserved
x2
– 4x + 6x – 24
= x(x – 4) + 6(x – 4)
= (x – 4)(x + 6)

18. ALGEBRIC IDENTITIES

19. THANK YOU