POLYNOMIALS ( TEST )

POLYNOMIALS

LESSON-12, DAY 5

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All the best 👍😊

POLYNOMIALS (LECTURE 11)

LESSON 11 POLYNOMIALS

21ST MAY

REVISION -POLYNOMIALS

GOOD MORNING!!

YESTERDAY,WE COMPLETED THE CHAPTER POLYNOMIALS !!

TODAY WE WILL BE GOING THROUGH THE TOPIC AGAIN AND SOLVE SOME IMPORTANT QUESTIONS!!

MEETING ID👇👇👇  

at 11:45 AM

QUESTIONS ARE GIVEN IN THE FORM OF A POWER POINT PRESENTATION

YOU CAN CLICK ON THE' <' OR' >' KEY AT THE BOTTOM OF IT TO VIEW THE 

OTHER SLIDES

MEETING ID👇👇👇

meet.google.com/hhb-utmk-jti

THERE WILL BE  A TEST ON POLYNOMIALS TOMORROW!

TAKE CARE AND KEEP SAFE!

POLYNOMIALS (LECTURE 10)

BLOG polynomials q 11 to q 16- 20TH MAY

GOOD MORNING BOYS ,

LEARNING OUTCOMES
BY THE END OF THIS LESSON YOU WILL BE TO UNDERSTAND AND APPLY STANDARD IDENTITIES TO

(A) FACTORISE THE GIVEN ALGEBRAIC EXPRESSION

(B)  EXPAND ( IN THE PRODUCT FORM )THE GIVEN ALGEBRAIC EXPRESSION
(C) VERIFY THE IDENTITY 
(D) FINDING THE POSSIBLE EXPRESSIONS FOR THE DIMENSIONS OF THE RECTANGLES AND THE CUBOIDS 

      CLICK HERE TO WATCH A VIDEO ON THE RECALL OF ALL THE IDENTITIES USING BINOMIAL  THEOREM 

              NCERT QUESTION  11  Pg 19

               NCERT QUESTION  12  Pg 19

             NCERT QUESTION  13  Pg 19

              NCERT QUESTION  14  Pg 19

              NCERT QUESTION  15  Pg 19

              NCERT QUESTION  16  Pg 19

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POLYNOMIALS (LECTURE 9)

POLYNOMIALS ex 2.5 polynomial q .5 - q 10 -19TH MAY

 GOOD MORNING BOYS ,

LEARNING OUTCOMES
BY THE END OF THIS LESSON YOU WILL BE TO UNDERSTAND AND APPLY STANDARD IDENTITIES TO

(A) FACTORISE THE GIVEN ALGEBRAIC EXPRESSION

(B)  EXPAND ( IN THE PRODUCT FORM )THE GIVEN ALGEBRAIC EXPRESSION

  CLICK HERE TO WATCH A VIDEO ON The EIGHT IMPORTANT ALGEBRAIC IDENTITIES 

              NCERT QUESTION  5  Pg 19
USING THE FOLLOWING IDENTITY ,FACTORISE 

NCERT QUESTION  6 Pg 19 

NCERT QUESTION  7 Pg 19 

NCERT QUESTION  8 Pg 19 

NCERT QUESTION  9 Pg 19

VERIFY THAT  

 NCERT QUESTION  10 Pg 19 

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POLYNOMIALS ( LECTURE 8)

Polynomials  (Lecture 8)


GOOD MORNING EVERYONE!!!!!!!!!!!!

From your earlier classes, you may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it.

You have  studied the following algebraic identities in earlier classes:
Write down these identities in your register

Today's learning outcome : How to apply these algebraic identities to factorize the algebraic 

expressions and also their utility in computations.

Let us solve Exercise 2.5

Do all these sums in your register

That's all for today. Tomorrow will meet with more sums, till then BYE

THANK YOU ALL 

POLYNOMIALS ( LECTURE 7)

LESSON 7 POLYNOMIALS

GOOD MORNING STUDENTS



yesterdays ' learning outcomes were:


I will be able to factorize polynomials by using

Todays learning outcomes  I will be able to 
 find the value of "k" 
factorise  a quadratic polynomial (splitting the middle term)
factorise a cubic polynomial

AQAD
1) Find the remainder  when x4+x3-2x2+x+1 is divided by x-1
 a)1
b)5
c)2
d)3



FEW INSTRUCTIONS
 The content in BLUE have to be written in the register as cw.

NOW LETS SOLVE Question 3.(EX 2.4)
Find the value of k, if x- 1 is a factor of p(x) in each of the following cases:

(ii)$p\left(x\right)=2x^2+kx+\sqrt{2}$
$p(x)=k{x}^2-\sqrt{2}$
(iv)$p\left(x\right)=k{x}^2-3x+k$
Solution LETS  SOLVE (ii) 

(ii) If x- 1 is a factor of polynomial 

$p\left(x\right)=2x^2+kx+\sqrt{2}$, then (USING FACTOR THEOREM) 
$p\left(1\right)=0$
$2(1{)}^2+k(1)+\sqrt{2}=0$
$k=-2-\sqrt{2}=-(2+\sqrt{2})$
So, value of k is $-(2+\sqrt{2})$.


(iv) If x- 1 is a factor of polynomial 
$p\left(x\right)=k{x}^2-3x+k$, then
$p\left(1\right)=0$
$k(1{)}^2+3(1)+k=0$
$k-3+k=0$
$k=\frac{3}{2}$
So, value of k is $\frac{3}{2}$

 Lets recall factorising quadratic polynomial by splitting the middle term


 .CLICK THE GIVEN LINK FOR  FACTORISATION OF QUADRATIC POLYNOMIAL

factorise quadratic polynomial


 LETS TAKE AN  EXAMPLE  ( TO  FACTORISE)  
Example 1 : 4x2 + 12x + 5
Solution: We have to find two numbers, whose
sum is 12 (middle term) and when
multiplied (first and last term) i.e. 4 X 5 we get 20

So these two numbers are -
10 + 2 = 12
10 X 2 = 20

Now we put both 10 and 2 numbers in the middle term of 12x and we get

∴ 4x2 + 12x + 5
 = 4x2 + 10x + 2x + 5
or 2x(2x + 5) + 1(2x + 5)
or (2x + 5)(2x + 1)

Try       4x2 + 8x - 5  (CW)         Ans ((2x + 5)(2x - 1)


solve Q3 of ex 2.4
Question 4
Factorise:
(i) $12x^2+7x+1$

lets solve 
Solution
Here we would be using splitting the middle term to factorise the polynomial
To factorise $a{x}^2+bx+c$, we should write b as the sum of two numbers whose product is ac
(i) $12x^2+7x+1$
Here a=12, c=1 and b=7 So $b=7=3+4$ ,$ac=12\times 1=12=3\times 4$
$12x^2+7x+1$
$=12x^2+4x+3x+1$
$=4x(3x+1)+1(3x+1)$
$=(3x+1)(4x+1)$


(iv) $3x^2-x-4$
Here a=3, c=-4 and b=-1 So, $b=-1=-4+3$, $ac=3\times (-4)=-12=(-4)\times 3$
$=3x^2-4x+3x-4$
$=x(3x-4)+1(3x-4)$
$=(3x-4)(x+1)$
  Lets see the video how to factorise cubic polynomial. click 👇 



$\mathrm{factorising\; cubic\; polynomialQ5}$
Factorise:

(i) $x^3-2x^2-x+2$

Solution
These are cubic polynomials and can be factorized using combination of long division method , remainder theorem and split middle term method
(i) Let $p\left(x\right)=x^3-2x^2-x+2$

STEP 1       Factors of 2 are ±1 and ± 2

STEP 2       By trial method, we find that

$p(-1)=(-1{)}^3-2(-1{)}^2-(-1)+2=0$(
$P(-1)=(-1{)}^3-2(-1{)}^2-(-1)+2=0$
STEP 3:-   So,(x+1) is factor of p(x)

. STEP 4  :-Now using the Long division method , we can find the quotient as

STEP 5 :-  Now, $Dividend=Divisor\times Quotient+Remainder$
$x^3-2x^2-x+2=(x+1)(x^2-3x+2)$

Factorizing the second part by split middle term method
$=(x+1)(x^2-x-2x+2)$
$=(x+1)\left[x\right(x-1)-2(x-1\left)\right]$
$=(x+1)(x-1)(x+2)$

Q5 (ii) $x^3-3x^2-9x-5$
LETS  SEE( ii) PART TOO 



(ii) Let 
$p\left(x\right)=x^3-3x^2-9x-5$
Factors of 5 are ±1 and ±5
By trial method, we find that
$p\left(5\right)=5^3-3\times 5^2-9\times 5-5=0$
So,(x-5) is factor of p(x)
Now using the Long division method , we can find the quotient as

Now, $Dividend=Divisor\times Quotient+Remainder$
$x^3-3x^2-9x-5=(x-5)(x^2+2x+1)$
Factorizing the second part by split middle term method
$=(x-5)(x^2+x+x+1)$
$=(x-5)x(x+1)+1(x+1)$
$=(x-5)(x+1)(x+1)$

HOME WORK  remaining parts 
(iii) $x^3+13x^2+32x+20$


(iv) $2y^3+y^2-2y-1$

I would encourage the students to try  and then check your answer.


(iii) Let $p\left(x\right)=x^3+13x^2+32x+20$
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
$p(-1)=(-1{)}^3+13(-1{)}^2+32(-1)+20=0$
So,(x+1) is factor ofp(x)
Now using the Long division method , we can find the quotient as

Now, Dividend = Divisor \times Quotient+ Remainder
$x^3+13x^2+32x+20=(x+1)(x^2+12x+20)$
Factorizing the second part by split middle term method
$=(x+1)(x^2+2x+10x+20)$
$=(x-5)\left[x\right(x+2)+10(x+2\left)\right]$
$=(x-5)(x+2)(x+10)$

(iv) Let $p\left(y\right)=2y^3+y^2-2y-1$
Factors of ab = 2 \times (-1) = -2 are ±1 and ±2
By trial method, we find that
$p\left(1\right)=2(1{)}^3+(1{)}^2-2\left(1\right)-1=0$
So,(y-1) is factor of p(y)
Now using the Long division method , we can find the quotient as

Now, $Dividend=Divisor\times Quotient+Remainder$
$\mathrm{Dividend=Divisor\times Quotient+Remainder2}{y}^3+y^2-2y-1=(y-1)(2y^2+3y+1)$
Factorizing the second part by split middle term method
$=(y-1)(2y^2+2y+y+1)$
$=(y-1)\left[2y\right(y+1)+1(y+1\left)\right]$
$=(y-1)(2y+1)(y+1)$

 HOMEWORK  
I would encourage the students to try  and then check your answer.

Q3,Q4 (remaining parts )
Factorize the  following

1. 3x3 –x2-3x+1
2. x3-23x2+142x-12

Answer
a) )(3x-1)(x-1)(x+1)
b)(x-1)(x-10)(x-12)

   AQAD     Solution ( c)


Q3 solution
(i) If x- 1 is a factor of polynomial $p\left(x\right)=x^2+x+k$, then
$p\left(1\right)=0$
$(1{)}^2+1+k=0$
$2+k=0$
$k=-2$

So, value of k is -2.
(iii) If x- 1 is a factor of polynomial $p\left(x\right)=k{x}^2-\sqrt{2}x+1$, then
$p\left(1\right)=0$
$k(1{)}^2-\sqrt{2}(1)+1=0$
$k=\sqrt{2}-1$

So, value of k is $\sqrt{2}-1$
$\mathrm{Q4\; (ii\; )2}{x}^2+7x+3$
Here a=2, c=3 and b=7 So $b=7=6+1$ ,$ac=2\times 3=6=6\times 1$
$=2x^2+6x+x+3$
$=2x(x+3)+1(x+3)$
$=(x+3)(2x+1)$

(iii) $6x^2+5x-6$
Here a=6, c=-6 and b=5 So, $b=5=9+(-4)$ ,$ac=6\times (-6)=-36=9\times (-4)$
$=6x^2+9x-4x-6$
$=3x(2x+3)-2(2x+3)=$