Problems on Remainder theorem

If  2x³ – (a – b)x² –(4b – 1)x + 4a is divisible by x² - 3x -28, then find the value of “a” and “b”.

Answer:

Factorizing x² - 3x -28 we get,

= x² - 7x + 4x -28

= x(x - 7) + 4(x - 7)

= (x- 7)(x + 4)

Let F(x) = 2x³ – (a – b)x² –(4b – 1)x + 4a

Now according to remainder theorem,  F(7) = 0 and F(-4) = 0 for complete divisibility.

Substituting x = 7 we get

686 –49 (a – b) – 7(4b – 1) + 4a = 0

Or,  693 – 45a + 21b = 0

Or, 45a – 21b = 693    -------------(1)

Similarly substituting x = 4, we get

-128 – 16(a – b) + 4(4b – 1) + 4a = 0

Or, -132 – 12a + 32b = 0

Or, 12a – 32b = -132  ---------(2)

Solving Equation 1 and 2 we get,

“a” = 21

“b” = 12

Problems on Remainder theorem

Show that x⁵ – 7x³ – 12x + 18 is divisible by x² + 2x -3

Answer:

Factorizing x² + 2x -3 we get,

= x² + 3x - x -3

= x(x + 3) -1(x +3)

= (x+ 3)(x -1)

Now according to remainder theorem, if F(x) = x⁵ – 7x³ – 12x + 18

Then, F(-3) = 0 and F(1) = 0 for complete divisibility.

Substituting x = -3 we get F(-3) = 0. Similarly for x = 1, we get F(1) = 0.

So, x⁵ – 7x³ – 12x + 18 is divisible by x² + 2x - 3.

If  x² - 2ax + 15 is divisible by x + 5, then find the value of “a”

Answer:

Now according to remainder theorem, if F(x) = x² - 2ax + 15

Then, F(-5) = 0 for complete divisibility.

But F(-5) = 25 + 10a +  15  = 40 + 10a

Since F(-5) = 0, we get 40 + 10a = 0

Or, a = -4

 

multiplication by use of detached coefficient

Multiply (x - 3x² +x³) (x +2x³-x⁵ ) as far as 4^th power of x.

multiplication by use of detached coefficient

Multiply using the method of detached coefficient

Multiply (1 – x +2x² +x³) by (1 +x- 2x³ + x⁴) as far as 4^th power of x.

Application of detached coefficient in multiplication

Application  of detached coefficient  in multiplication


Example: Expand (2 – x + 3x³ – x⁴) by (1 – 2x² + x³ + 2x⁵) up to the third power of X.

Algebric multiplication using detached coeffficient

Use of detached coefficients in multiplication

Factorisation

Resolve into factors:

2 + m – m²

Answer:

= 2 +2m - m – m²

=2(1 + m) – m(1 + m)

= (2 – m)(1 + m)

Resolve into factors:

98 – 7y – y²

Answer:

98 – 7y – y²

= 98 – 14y + 7y – y²

= 14(7 – y) + y(7 – y)

= (14 + y)(7 – y)

Resolve into Factors:

y²- 78 + 7y 

Answer:

Y²-78 + 7y

= y²+ 7y -78

= y²+ 13y – 6y -78

= y(y +13) – 6(y + 13)

= (y + 13)(y – 6)

How to solve a magic square STD IV

HOW TO SOLVE PROBLEMS ON MAGIC SQUARE

What is a magic square?

Please see the square below:

8	15	7
9	10	11
13	5	12

 

In this square the sum of all the rows, columns and diagonals add up to 30. In a magic square the sum of the rows columns and diagonals always add up to give the same value.

What is the magic?

Rule 1:  the central number (i.e. 10 in the magic square above) multiplied by 3 is the sum of the rows or columns or diagonals.

Rule 2: The central number is always half of the adjacent two numbers, for example (8 + 12)/ 2 = 10,

(15 + 5)/2 = 10, (9 + 11)/2 = 10 or (7 + 13)/ 2 = 10

So, how do we proceed?

Let us take the above example;

9		11
13

      Can you solve this? It is so easy,

      the central number is (9 + 11)/2 = 10

      So, the sum of rows or columns or diagonals = 10 x 3 = 30

Now I am sure you can complete the table.

Let us look at other combinations possible:

		7
	10
		12

In this case since the central number is given, we immediately have the

Sum of row or column or diagonal = 10 x 3 = 30

Once you know this, rest is easy to fill up.

9
13	5

How about this combination?

Finding the central number is difficult!!

You may choose to use this formula to find the central number:

[(9 + 5) + 2 x 13]/4 = 40/4 = 10

Sum of row or column or diagonal = 10 x 3 = 30

Once you know this, rest is easy to fill up.

Now do you feel any better with magic squares?