*We at St Columba's School, are ready to take the next step - and move towards using Google Classroom as a tool to enhance the teaching learning environment. *This shift involves the school giving each student a unique email address, which he will need to setup ( the teachers will assist him in doing the same). This email will facilitate each student to interact with his teachers. * The student should use this email ID only for: -interacting with teachers on school /subject related matters -keep the words and expressions as he would- if he were interacting with us as a Columban student -use the facilities connected to this email ID only for google meets organised with the permission of a teacher -he should have a copy of the permission given ( when and by which teacher and for what purpose) -do not access any of the other facilities connected with this email address unless he first seeks the permission of a teacher (He should have a copy of the permission given - when and by which teacher and for what purpose). *Please note that Google and St Columba's have legal obligations by giving you access to this email address and hence, you will need to sign an agreement when creating this email ID. *Be informed that the history of use of this ID shall have a digital footprint !
Let us go through the guidelines for the blog once again:
·Red, is to be 🖉 in your 📖
·Blue is to be 👀 by ➤
·Green is to be 💬🖉📗 for home work
take your SET-A Mathematics 📖
Our📃. It will be 👍, if you use good presentation and cursive 📃
Make a column on the RHS, if you need to do any rough work
Leave ⓶ lines where you finished yesterday’s work and draw a horizontal line
Write today's 📆
LEARNING OUTCOMES Covered So far:
1. RECALL WHAT IS AN ALGEBRAIC EXPRESSION AND DEFINE POLYNOMIAL.
2. RECALL AND DEFINE TERMS AND COEFFICIENTS.
3. DEFINE DEGREE OF A POLYNOMIAL
4. CLASSIFY THE POLYNOMIALS ON THE BASIS OF NUMBER OF TERMS AND DEGREE.
5. COMPREHEND AND MEMORIZE ABOUT SOME SPECIAL POLYNOMIALS. 6. EVALUATE VALUE OF A POLYNOMIAL
7 .EVALUATE ZERO OF A POLYNOMIAL.
8. Know that a polynomial can be ÷ by another polynomial.
9.Apply division process to polynomials.
10. Comprehend remainder theorem.
TODAY'S LEARNING OUTCOMES:
I WILL BE ABLE TO:
apply remainder theorem to calculate the remainder when a polynomial is ÷ by another polynomial.
Dear Students, a worksheet on Lines & Angles has been sent through redox assignment tab to you. Please download using your student login on redox app or school website. Complete it by 18th May and then you may discuss the queries with me.
In the questions given below p(x) is divide by g(x). Q1 & Q5 are done as sample for you as an application of remainder theorem. Solve the other questions as per the steps given in the table👇
Let us go through the guidelines for the blog once again:
·Red, is to be 🖉 in your 📖
·Blue is to be 👀 by ➤
·Green is to be 💬🖉📗 for home work
take your SET-A Mathematics 📖
Our📃. It will be 👍, if you use good presentation and cursive 📃
Make a column on the RHS, if you need to do any rough work
Leave ⓶ lines where you finished yesterday’s work and draw a horizontal line
Write today's 📆
LEARNING OUTCOMES Covered So far:
1. RECALL WHAT IS AN ALGEBRAIC EXPRESSION AND DEFINE POLYNOMIAL.
2. RECALL AND DEFINE TERMS AND COEFFICIENTS.
3. DEFINE DEGREE OF A POLYNOMIAL
4. CLASSIFY THE POLYNOMIALS ON THE BASIS OF NUMBER OF TERMS AND DEGREE.
5. COMPREHEND AND MEMORIZE ABOUT SOME SPECIAL POLYNOMIALS. 6. EVALUATE VALUE OF A POLYNOMIAL
7 .EVALUATE ZERO OF A POLYNOMIAL.
TODAY'S LEARNING OUTCOMES:
I WILL BE ABLE TO: 1) know that a polynomial can be ÷ by another polynomial. 2) apply division process to polynomials. 3) comprehend remainder theorem. 4) apply remainder theorem to calculate the remainder when a polynomial is ÷ by another polynomial.
Q1) 10 - 6 = 4 , here 4 is known as a (remainder / sum)
Q2) 10 ÷ 2 = 5,here 5 is known as a (remainder/ quotient)
Q3) (10x - 1) - (3) = (10x -4), here (10x -4) is a _____
Q4) (2x + 4) ÷ 2 = POSSIBLE ? , please watch the following video (for just first 4min 45 seconds) to check
please observe the image given below
Q5) Divide (2x² + 4x - 7) ÷ (x + 1) A5) STEP-1: Write in division format STEP-2: Divide 2x² by x {1st term of dividend by 1st term of divisor} STEP-3: Multiply (x + 1) by the answer from step-2 STEP-4: Subtract & bring down -7{This forms the new dividend} STEP-5: Divide 2x by x {1st term of new dividend by 1st term of divisor}
STEP-6: Multiply (x + 1) by the answer from step-5 STEP-7: Subtract. The remainder is ____
The above process
gives us two OUTPUTS:
FIRST: The quotient
SECOND: The remainder Please ➤👇
You heard this word "tedious", in the above audio.
Please write this new word in your diary with today's date:
TEDIOUS : means , too long, slow or dull
Learn to pronounce
Cleaning of our rivers such as Yamuna & Ganga
had become a tedious process,
because,
we as the citizens of India
were not making full contributions to the process.
In the questions given below p(x) is divide by g(x). Q8 & Q12 are done as sample for you as an application of remainder theorem. Solve the other questions as per the steps given in the table👇
Text in red has to be noted down as its your class work. Text in blue are videos (watch )
Text in green is your homework
AQAQ
The polynomial of type ax2 + bx + c, a = 0 is of type
(a) linear
(b) quadratic
(c) cubic
(d) Biquadratic
In the previous class we discussed about
1. CLASSIFY THE POLYNOMIALS ON THE BASIS OF NUMBER OF TERMS AND DEGREE. 2. COMPREHEND AND MEMORIZE ABOUT SOME SPECIAL POLYNOMIALS.
TODAY'S LEARNING OUTCOMES: I WILL BE ABLE TO:
1. EVALUATE VALUE OF A POLYNOMIAL 2.EVALUATE ZERO OF A POLYNOMIAL.
Value of a Polynomial For Example: Find value of this polynomialp(x) = x + 2 at x =1
.(step i) Consider p(x) = x +1 .
,( step ii) If we put x = 1 in p(x), we get (step (iii) p(1) = 1 + 2 = 3
Thus,3 is the value of the polynomial p(x). at x = 1 Zero of a Polynomial : The value of variable for which the polynomial becomes zero is called as the zero of the polynomial . For Example: Find zero of this polynomial p(x) = x + 2. (stepi) p(x) = x + 2. (stepii) If we put x = -2 in p(x), , we get .(stepiii) p(-2) = -2 + 2 = 0 Thus,( -2) is a zero of the polynomial p(x)................................ why?. (NOTE :- The value of polynomial p(x) has become zero , when x=(_-2 ). TO KNOW MORE ,CLICK THE LINK GIVEN BELOW
Lets solve another Example: (Read carefully and try to understand )
Find value of polynomial 3a2 + 5a + 1 at a = 3
(i) Here, p(a) = 3a2 + 5a + 2. (ii) Now, substituting a = 3, we get ,(iii) p(3) = 3 x (3)2 + 5 x 3 + 2 = 27 + 15 + 2 = 44
. Thus the value of polynomial p(x) is 44, if a = 3
Find zero of polynomial 3a2 + 5a + 1 .
(i) Here, p(a) = 3a2 + 5a + 2 .(ii) Now, substituting a = ( -1), we get, (iii) p(-1 ) = 3 x (-1 )2 + 5 x (-1) + 2 = 3 - 5 + 2 = 0
. Thus the zero of polynomial p(x) is (-1)
so Answer these questions
The value of the polynomial p(x) = 2x + 5 is,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,( 5 or 7 ) , at x = 1,, The value of p (x) = x + 2 is 2 , if the value of x is ..............................( one / zero) The value of p (x) = x2 + 4x + 2 at x = ( -1) is ................. ( -1 or 7 )
The zero of the polynomial p(x) = 2x + 5 is,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,( 5/2 or -5./2 ),, The number of zeros of x2 + 4x + 2 is...............................( one or two)
Q4 (i)
Let's find the zero of given polynomial Q4 (vi)p(x) = ax , a ≠ 0 since , p(x) = 0 ⇒ ax = 0 ⇒ x = 0 ,Thus zero of p(x) is 0
(vii) p(x) = cx +d
since , p(x) = 0 ⇒ cx +d = 0 ⇒ cx = -d ⇒ x = -d/c, Thus zero of p(x) is -d / c
NOTE -:
(i) A non-zero constant polynomial has no zero .(ii) A linear polynomial has one and only one zero .(iii) A zero of a polynomial might not be 0 or 0 might be a zero of a polynomial .(iv) A polynomial can have more than one zero. (quadratic , cubic.......)
Q1.Check whether at x = -1/7 is zero of the polynomial p(x) = 7x + 1. Q2.Find zero of the polynomial p(x) = 2x+ 2.
Example 2
From text book
Now we are able to solve EXERCISE 2.4
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