POLYNOMIALS (LECTURE 6)

LESSON 6 POLYNOMIALS

GOOD MORNING!!
Today's learning outcomes:
I WILL BE ABLE TO;
1. divide one polynomial by the other
2. recall and define division algorithm for polynomials
3. apply division algorithm to solve questions.


Before you join the meet ,plz go through the blog once.

Please join the google meet classroom by clicking on the following link. I will be available here from 8:00 - 9: 00 am to sort your queries.

Rules:

• Keep your mic on mute. You will switch it on only to ask questions or answer the questions when called on to do so. 
• Enter meeting with your own names. 
• Please join in on time as late entries will not be entertained. 
• Keep your videos off.

                     READ ,VIEW AND UNDERSTAND



WHAT IS DIVISION ALGORITHM?

LET'S START WITH AN EG OF DIVISION OF TWO NUMBERS


11 ÷ 2

As you can see in above eg.

11 can be written as 2 x5 +1

or 11 =2 x5 +1
or DIVIDEND = DIVISOR X QUOTIENT + REMAINDER
we can have remainder =0 or will be less than divisor

WHERE WE LEARNT ABOUT  DIVISION ALGORITHM IN CLASS 10??

IS IT TRUE FOR POLYNOMIALS ALSO?


FOR THIS FIRST , WE WILL REVISE THE PROCESS OF DIVISION

WATCH  THIS VIDEO☟

HAVE U SEEN THE VIDEO?

NOW WE WILL SEE IF THE DIVISION ALGORITHM IS TRUE FOR POLYNOMIALS OR NOT


WATCH THIS ☟

 WHAT DID YOU LEARN?


                                                       CLASS WORK 

DIVISION ALGORITHM FOR POLYNOMIALS(NOTE DOWN)


If p(x) and g(x) are any two polynomials with g(x) ≠0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).
This result is known as the Division Algorithm for polynomials

✴✴.STEPS TO PERFORM THE DIVISION OF POLYNOMIALS (FOR THIS WE WILL CONSIDER EX 2.3 Q1 (ii))

Q1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following : 
(ii) p(x) = x⁴ – 3x² + 4x + 5,  g(x) = x² + 1 – x 

STEP 1. arrange the terms of the dividend and the divisor in the decreasing order of their degrees. (STANDARD FORM)

SO g(x) = x²  – x +1

STEP 2. IF any term is missing in dividend then we insert that term with 0 as a coefficient
here, x³ is missing in p(x) so we will insert 0x³

p(x) = x⁴ + 0x³  – 3x² + 4x + 5

now, we will divide p(x) by g(x) as shown below

⭐⭐we will perform division till the remainder becomes 0 or degree of remainer is less than degree of divisor
like here, deg of remainer is 0 (why??)
 ✸you have learnt in class ix about degree of constant polynomial  )


degree of divisor is 2


Q2  Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: 

❄❄ we can check this by looking at the remainder after division process
             If it is 0 then it is a factor


(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12 

Let p(x)=  t² – 3
g(x)=  2t⁴ + 3t³ – 2t² – 9t – 12


🔴 Both are in standard form

🔴 since remainder is 0, p(x) is a  factor of g(x)


(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2 
Let p(x)= 3x⁴ + 5x³ – 7x² + 2x + 2 

g(x) = x² + 3x + 1

SINCE REMAINDER IS 0 ,g(x) is factor


4. On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).

let p(x) =  x³ – 3x² + x + 2

g(x) =?(divisor)

q(x) =  x – 2(quotient)
r(x) =  -2x +4 (remainder)




❄❄ 
since,DIVIDEND = DIVISOR X QUOTIENT + REMAINDER

 SO,        DIVISOR =( DIVIDEND -REMAINDER )/QUOTIENT

 NOW ,DIVIDEND -REMAINDER= x³ – 3x² + x + 2-( -2x +4 )
= x³ – 3x² + 3x  -2

In next step we will divide x³ – 3x² + 3x  -2   by  x – 2

5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x) 

deg p(x) (dividend )  will be same as deg q(x) (quotient)

if  divisor (g (x)) is a constant.

eg. p(x) = 6x ² +2x +2

       g(x) = 2

    then, q(x) = 3x ² +x +1 and  r(x) =0

🔴YOU CAN VERIFY THIS BY APPLYING DIVISION ALGORITHM

                             HOME WORK

EX 2.3

Q1 (I) (III)

Q2(III)

Q5(II)(III)

WE END OUR CLASS HERE!!

PLZ FEEL FREE TO ASK IN CASE OF ANY QUERY!!

TAKE CARE!!