Showing posts with label (012) GRE Math Test - Important Key points and formulas. Show all posts
Showing posts with label (012) GRE Math Test - Important Key points and formulas. Show all posts

Saturday, December 1, 2018

107-GRE Math-9-Important Key points and formulas

In continuation of part - 8, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued):

3) Synthetic Division: 

Example-4: Divide 4x4+(13/3)x3-(23/3)x2+11x-(18/3) by (x-2/3) using the synthetic division method.

Solution:

1) Write all the terms in descending order:
     4x4+(13/3)x3-(23/3)x2+11x-(18/3)
2) Write this polynomial in the coefficient form:
     [4, 13/3, -23/3, 11, -18/3]
3) Now we will see the division using the synthetic division method.
     a) Here the divisor is (x-2/3) so we have x-2/3=0, x=2/3.
b) Here the quotient is 4x3+7x2-3x+9 and the remainder is 0.
c) So we have 4x4+(13/3)x3-(23/3)x2+11x-(18/3) = (x-2/3) (4x3+7x2-3x+9)+0.

Example-5: Divide 3x5+(9/5)x4-2x-(6/5) by (x+3/) using synthetic division method.


Solution:
1) Write all the terms in descending order:
     3x5+(9/5)x4+0x3+0x2-2x-(6/5)
2) Write this polynomial in the coefficient form:
     [3, 9/5, 0, 0, -2, 6/5]
3) Now we will see the division using the synthetic division method.
     a) Here the divisor is (x+(3/5)) so we have x+3/5=0, x=-3/5.

b) Here the quotient is 3x4-2 and the remainder is 0.
c) So we have 3x5+(9/5)x4-2x-(6/5) = (x+3/5) (3x4-2)+0.

In the next part, we will see a few examples and some essential formulae.

ANIL SATPUTE

Friday, July 6, 2018

105-GRE Math-8-Important Key points and formulas

In continuation of part - 7, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued):

3) Synthetic Division: 

Example 2: Divide 4x6-20x5+7x2-46x+55 by (x-5) using the synthetic division method.

Solution:

1) Write all the terms in descending order:
     4x6-20x5+0x4+0x3+7x2-46x+55
2) Write this polynomial in the coefficient form:
     [4, -20, 0, 0, 7, -46, 55]
3) Now we will see the division using the synthetic division method.
     a) Here the divisor is (x-5) so we have x-5=0, x=5.
   b) Here the quotient is 4x6-20x5+0x4+0x3+7x2-46x+55 which can also be written as 4x6-205+7x2-46x+55. Here the remainder is 0.
    c) So we have 4x6-205+7x2-46x+55 = (x-5) (4x5+7x-11)+0.

Example 3: Divide 5x5+20x4-2x3-8x2+3x+15 by (x+4) using the synthetic division method.


Solution:
1) Write all the terms in descending order:
     5x5+20x4-2x3-8x2+3x+15
2) Write this polynomial in the coefficient form:
     [5, 20, -2, -8, 3, 15]
3) Now we will see the division using the synthetic division method.
     a) Here the divisor is (x+4) so we have x+4=0, x=-4.

   b) Here the quotient is 5x5+20x4-2x3-8x2+3x+15. Here the remainder is 3.
   c) So we have 5x5+20x4-2x3-8x2+3x+15 = (x+4) (5x4-2x2+3)+3.

In the next part, we will see a few examples and some essential formulae.

Friday, June 22, 2018

104-GRE Math-7-Important Key points and formulas

In continuation of part - 6, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued):

1) Factor Theorem: 

For a polynomial p(x), we say that (x-a) is a factor of the polynomial p(x), if we divide polynomial p(x) by (x-a), then we get the remainder as 0. i.e. p(x) = q(x) (x-a) + 0 where q(x) another polynomial called quotient (a result obtained by dividing p(x) by (x-a)).

2) Division formula in algorithm form: 

In the algorithm formula of division,  p(x) = q(x) d(x) + r(x), 
p(x) is polynomial
q(x) is quotient
d(x) is divisor
r(x) is remainder

Any polynomial can be written in the form "p(x) = q(x) d(x) + r(x)"

3) Synthetic Division: 

A few important points are to be noted to learn synthetic division.

       a) A coefficient is a real number which is the multiplicative factor of the term in the polynomial.  
Example:
1) Let the polynomial be 8x4-3x2+7x+3. First, write all the terms in descending order. So we write it as 8x4+0x3+3x2+7x+3. Now we write it in coefficient form as 8, 0, -3, 7, 3. Here, as the power of the polynomial is 4, the total number of terms of the polynomial will be 5. 
2) Let the polynomial be 7x5First, write all the terms in descending order. So we write it as 7x5+0x4+0x3+0x2+0x+0. Now we write it in coefficient form as 7, 0, 0, 0, 0, 0. Here, as the power of the polynomial is 5, the total number of terms of the polynomial will be 6. 
         b) In the case of synthetic division, the divisor must be of the first degree of the form (x - a). Example (x - a) = 0, so the divisor is (x - a) with x = a.

Example: Divide 3x6-6x5+7x4-18x3+13x2-9x-2 by (x-2) using the synthetic division method.

Solution:
1) Write all the terms in descending order:
     3x6-6x5+7x4-18x3+13x2-9x-2

2) Write this polynomial in the coefficient form:
     [3, -6, 7, -18, 13, -9, -2]

3) Now we will see the division using the synthetic division method.

     a) Here the divisor is (x-2) so we have x-2=0, x=2.
   b) Here the quotient is 3x5+0x4+7x3-4x2+x+2 which can also be written as 3x5+7x3-4x2+x+2. here the remainder is 0.
     c) So we have 3x6-6x5+7x4-18x3+13x2-9x-2 = (x-2) (3x5+7x3-4x2+x+2)+0.

In the next part, we will see a few examples and some essential formulae.

Friday, August 11, 2017

101-GRE Math-6-Important Key points and formulas

In continuation of part - 5, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued)

4) (x - a) (x - b) = x (x - b) - a (x - b)
                         = x 2 - b x - a x + ab
                         = x 2 - (a + b) x + ab
Generally, we call x 2 as the first term, (a + b) x as the middle term, and ab as the last term. 

Basic concept: 

a) Step-1: See the sign of the last term.
b) Step-2: Here it is "+" so factorize the product of the coefficient of the first term (here it is 1) and the last term in such a way that the SUM of these two factors must be the coefficient of the middle term.
c) Step-3: Get the factors.

Example-1:

Factorize: x 2 - 10 x + 21.

a) Step-1: Here sign of the last term 21 is "+"
b) Step-2: The coefficient of the first term is 1 and the last term is 21, so the product of 1 and 21 is 21. Now the factors of 21 are 3 and 7 and as the sign of the last term is "+", their addition is -3 - 7 = -10 which is the coefficient of the middle term.  
                         = x 2 - 10 x + 21
                         = x 2 - (3 + 7) x + (3 x 7)
                         = x 2 - 3 x - 7 x + (3 x 7)
                         = x (x - 3) - 7 (x - 3)
                         = (x - 3) (x - 7)
c) Step-3: So the factors of x 2 - 10 x + 21 are (x - 3) and (x - 7) 

Example-2:

Factorize: 8 x 2 - 14 x + 5.

a) Step-1: Here sign of the last term 5 is "+"
b) Step-2: The coefficient of the first term is 8 and the last term is 5, so the product of 8 and 5 is 8 X 5. Now the factors of 8 X 5 are 2, 2, 2, and 5 and as the sign of the last term is "+", so, we take two factors in such a way that their sum will be 14. Here 2, 2, 2, and 5 will give us 4 and 10. So, here addition is -4 - 10 = -14 which is the coefficient of the middle term.  
                         = 8 x 2 - 14 x + 5
                         = 8 x 2 - (4 + 10) x + 5
                         = 8 x 2 - 4 x - 10 x + 5
                         = 4 x (2 x - 1) - 5 (2 x - 1)
                         = (4 x - 5) (2 x - 1)
c) Step-3: So the factors of 8 x 2 - 14 x + 5 are (2 x - 1) and (4 x - 5).

In the next part, we will see a few examples and some essential formulae.

Thursday, August 10, 2017

100-GRE Math-5-Important Key points and formulas

In continuation of part - 4, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued)

3) (x - a) (x + b) = x (x + b) - a (x + b)
                          = x 2 + b x - a x - ab
                          = x 2 - (a - b) x - ab
Generally, we call x 2 as the first term, (a + b) x as the middle term, and ab as the last term. 

Basic concept: 

a) Step-1: See the sign of the last term.
b) Step-2: Here it is "-" so factorize the product of the coefficient of the first term (here it is 1) and the last term in such a way that the DIFFERENCE of these two factors must be the coefficient of the middle term.
c) Step-3: Get the factors.

Example-1:

Factorize: x 2 - 4 x - 21.

a) Step-1: Here sign of the last term 21 is "-"
b) Step-2: The coefficient of the first term is 1 and the last term is 21, so the product of 1 and 21 is 21. Now the factors of 21 are 3 and 7 and as the sign of the last term is "-", their subtraction is 3 - 7 = -4 which is the coefficient of the middle term. (Note: Here, the coefficient of the middle term is positive so we took it as 3 - 7).  
                         = x 2 - 4 x - 21
                         = x 2 + (3 - 7) x - (3 x 7)
                         = x 2 + 3 x - 7 x - (3 x 7)
                         = x (x + 3) - 7 (x + 3)
                         = (x - 7) (x + 3)
c) Step-3: So the factors of x 2 - 4 x - 21 are (x + 3) and (x - 7) 

Example-2:

Factorize: 8 x 2 - 18 x - 5.

a) Step-1: Here sign of the last term 5 is "-"
b) Step-2: The coefficient of the first term is 8 and the last term is 5, so the product of 8 and 5 is 8 X 5. Now the factors of 8 X 5 are 2, 2, 2, and 5 and as the sign of the last term is "-", so, we take the two factors in such a way that their difference will be 18. Here 2, 2, 2, and 5 will give us 2 and 20. So, here subtraction is 2 - 20 = -18 which is the coefficient of the middle term.  
                         = 8 x 2 - 18 x - 5
                         = 8 x 2 + (2 - 20) x - 5
                         = 8 x 2 + 2 x - 20 x - 5
                         = 2 x (4 x + 1) - 5 (4 x + 1)
                         = (2 x - 5) (4 x + 1) 
c) Step-3: So the factors of 8 x 2 - 18 x + 5 are (2 x - 5) and (4 x + 1).

In the next part, we will see the remaining 1 type in detail. These 1 types are given below.
4) (x - a) (x - b) = x 2 - (a + b) x + ab
In the next part, we will see a few examples and some essential formulae.

Wednesday, August 9, 2017

99-GRE Math-4-Important Key points and formulas

In continuation of part - 3, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued)

2) (x + a) (x - b) = x (x - b) + a (x - b)
                          = x 2 - b x + a x - ab
                          = x 2 + a x - b x - ab
                          = x 2 + (a - b) x - ab
Generally, we call x 2 as the first term, (a + b) x as the middle term, and ab as the last term. 

Basic concept: 

a) Step-1: See the sign of the last term.
b) Step-2: Here it is "-" so factorize the product of the coefficient of the first term (here it is 1) and the last term in such a way that the DIFFERENCE of these two factors must be the coefficient of the middle term.
c) Step-3: Get the factors.

Example-1:

Factorize: x 2 + 4 x - 21.

a) Step-1: Here sign of the last term 21 is "-"
b) Step-2: The coefficient of the first term is 1 and the last term is 21, so the product of 1 and 21 is 21. Now the factors of 21 are 3 and 7 and as the sign of the last term is "-", their subtraction is 7 - 3 = 4 which is the coefficient of the middle term. (Note: Here, the coefficient of the middle term is positive so we took it as 7 - 3).  
                         = x 2 + 4 x - 21
                         = x 2 + (7 - 3) x - (3 x 7)
                         = x 2 + 7 x - 3 x - (3 x 7)
                         = x (x + 7) - 3 (x + 7)
                         = (x - 3) (x + 7)
c) Step-3: So the factors of x 2 + 4 x - 21 are (x - 3) and (x + 7) 

Example-2:

Factorize: 8 x 2 + 18 x - 5.

a) Step-1: Here sign of the last term 5 is "-"
b) Step-2: The coefficient of the first term is 8 and the last term is 5, so the product of 8 and 5 is 8 X 5. Now the factors of 8 X 5 are 2, 2, 2, and 5 and as the sign of the last term is "-", so, we take the two factors in such a way that their difference will be 18. Here 2, 2, 2, and 5 will give us 2 and 20. So, here subtraction is 20 - 2 = 18 which is the coefficient of the middle term.  
                         = 8 x 2 + 18 x - 5
                         = 8 x 2 + (20 - 2) x - 5
                         = 8 x 2 + 20 x - 2 x - 5
                         = 4 x (2 x + 5) - 1 (2 x + 5)
                         = (2 x + 5) (4 x - 1) 
c) Step-3: So the factors of 8 x 2 + 18 x + 5 are (2 x + 5) and (4 x - 1).

In the next part, we will see the remaining 2 types in detail. These 2 types are given below.

3) (x - a) (x + b) = x 2 - (a - b) x - ab
4) (x - a) (x - b) = x 2 - (a + b) x + ab

In the next part, we will see a few examples and some essential formulae.

Tuesday, August 8, 2017

98-GRE Math-3- Important Key points and formulas

In continuation of part - 2, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

Factorization (Continued):

b) Factorization of Polynomial: 

Here following 4 possibilities can be studied to understand factorization in a better way.

1) (x + a) (x + b) = x (x + b) + a (x + b)
                          = x 2 + b x + a x + ab
                          = x 2 + a x + b x + ab
                          = x 2 + (a + b) x + ab
2) (x + a) (x - b) = x (x - b) + a (x - b)
                          = x 2 - b x + a x - ab
                          = x 2 + a x - b x - ab
                          = x 2 + (a - b) x - ab
3) (x - a) (x + b) = x (x + b) - a (x + b)
                          = x 2 + b x - a x - ab
                          = x 2 - a x + b x - ab
                          = x 2 - (a - b) x - ab

Note: Here formulae 2 and 3 are of the same type. the coefficient of the middle term is the difference of the Constance and the sign is to be taken from the greater Constance. 
4) (x - a) (x - b) = x (x - b) - a (x - b)
                          = x 2 - b x - a x + ab
                          = x 2 - a x - b x + ab
                          = x 2 - (a + b) x + ab

Now we will study these types in detail:
1) (x + a) (x + b) = x (x + b) + a (x + b)
                          = x 2 + b x + a x + ab
                          = x 2 + a x + b x + ab
                          = x 2 + (a + b) x + ab

Generally, we call x 2 as the first term, (a + b) x as the middle term, and ab as the last term. 

Basic concept: 

a) Step-1: See the sign of the last term.
b) Step-2: Here it is "+" so factorize the product of the coefficient of the first term (here it is 1) and the last term in such a way that the SUM of these two factors must be the coefficient of the middle term.
c) Step 3: Get the factors.

Example-1:

Factorize: x 2 + 10 x + 21.

a) Step 1: Here sign of the last term 21 is "+"
b) Step 2: The coefficient of the first term is 1 and the last term is 21, so the product of 1 and 21 is 21. Now the factors of 21 are 3 and 7 and as the sign of the last term is "+", their addition is 3 + 7 = 10 which is the coefficient of the middle term.
                         = x 2 + 10 x + 21
                         = x 2 + (3 + 7) x + (3 x 7)
                         = x 2 + 3 x + 7 x + (3 x 7)
                         = x (x + 3) + 7 (x + 3)
                         = (x + 3) (x + 7)
c) Step-3: So the factors of x 2 + 10 x + 21 are (x + 3) and (x + 7) 

Example-2:

Factorize: 8 x 2 + 14 x + 5.

a) Step 1: Here sign of the last term 5 is "+"
b) Step 2: The coefficient of the first term is 8 and the last term is 5, so the product of 8 and 5 is 8 X 5. Now the factors of 8 X 5 are 2, 2, 2, and 5 and as the sign of the last term is "+", so, we take two factors in such a way that their sum will be 14. Here 2, 2, 2, and 5 will give us 4 and 10. So, here addition is 4 + 10 = 14 which is the coefficient of the middle term.
                         = 8 x 2 + 14 x + 5
                         = 8 x 2 + (4 + 10) x + 5
                         = 8 x 2 + 4 x + 10 x + 5
                         = 4 x (2 x + 1) + 5 (2 x + 1)
                         = (4 x + 5) (2 x + 1)
c) Step 3: So the factors of 8 x 2 + 14 x + 5 are (2 x + 1) and (4 x + 5).
In the next part, we will see the remaining 3 types in detail. These 3 types are given below.
2) (x + a) (x - b) = x 2 + (a - b) x - ab
3) (x - a) (x + b) = x 2 - (a - b) x - ab
4) (x - a) (x - b) = x 2 - (a + b) x + ab  

In the next part, we will see a few examples and some essential formulae.