Math, asked by amritha855, 1 month ago

answer this question (class9)








no spamming ✖️​

Attachments:

Answers

Answered by VεnusVεronίcα
9

Without actual division, prove that 2x 5x³ + 2x² x + 2 is divisible by 3x + 2.

Let's split the middle term of x² – 3x + 2 firstly :

➺ x² – 3x + 2 = 0

➺ x² – x – 2x + 2 = 0

➺ x (x – 1) – 2 (x – 1) = 0

➺ (x – 1) (x – 2) = 0

x = 1 and x = 2

Let's now substitute these values in 2x⁴ – 5x³ + 2x² – x + 2 and verify :

Case 1 x = 1 :

➺ 2x⁴ – 5x³ + 2x² – x + 2 = 0

➺ 2 ( 1 )⁴ – 5 ( 1 )³ + 2 ( 1 )² – 1 + 2 = 0

➺ 2 ( 1 ) – 5 ( 1 ) + 2 ( 1 ) – 1 + 2 = 0

➺ 2 – 5 + 2 – 1 + 2 = 0

➺ 0 = 0

➺ LHS = RHS

Case 2 x = 2 :

➺ 2x⁴ – 5x³ + 2x² – x + 2 = 0

➺ 2 ( 2 )⁴ – 5 ( 2 )³ + 2 ( 2 )² – 2 + 2 = 0

➺ 2 ( 16 ) – 5 ( 8 ) + 2 ( 4 ) = 0

➺ 32 – 40 + 8 = 0

➺ 0 = 0

➺ LHS = RHS

The polynomial 2x⁴ – 5x³ + 2x² – x + 2 is divisible by the polynomial x² – 3x + 2.

_____________________

If a + b + c = 5 and ab + bc + ca = 10, then prove that + + 3abc = 25.

Squaring on both sides :

➺ (a + b + c)² = 5²

➺ a² + b² + c² + 2 (ab + bc + ca) = 25

➺ a² + b² + c² + 2 ( 10 ) = 25

➺ a² + b² + c² + 20 = 25

➺ a² + b² + c² = 25 – 20

a² + b² + c² = 5

We know the algebraic identity :

a³ + b³ + c³ – 3abc = (a + b + c) [a² + b² + c² – (ab + bc + ca)]

Let's substitute and verify :

➺ – 25 = ( 5 ) [5 – ( 10 ) ]

➺ – 25 = 5 ( – 5 )

➺ – 25 = – 25

Hence, verified.

_____________________

Without actual division, prove that 3x² 13x + 15 is exactly divisible by + 2x 3.

Let's now split the middle term of x² + 2x – 3 :

➺ x² + 2x – 3 = 0

➺ x² – x + 3x – 3 = 0

➺ x (x – 1) + 3 (x – 1) = 0

➺ (x – 1) (x + 3) = 0

x = 1 and x = 3

Now, let's substitute the values in x³ – 3x² – 13x + 15 and verify :

Case 1 x = 1 :

➺ x³ – 3x² – 13x + 15 = 0

➺ ( 1 )³ – 3 ( 1 )² – 13 + 15 = 0

➺ 1 – 3 ( 1 ) + 2 = 9

➺ 3 – 3 = 0

➺ 0 = 0

➺ LHS = RHS

Case 2 x = 3 :

➺ x³ – 3x² – 13x + 15 = 0

➺ (– 3)³ – 3 (– 3)² – 13 (– 3) + 15 = 0

➺ – 27 – 3 ( 9 ) + 39 + 15 = 0

➺ – 27 – 27 + 54 = 0

➺ – 54 + 54 = 0

➺ 0 = 0

➺ LHS = RHS

Therefore, x³ – 3x² – 13x + 15 is divisible by x² + 2x – 3.

_____________________

Find the value of a and b so that the polynomial 10x² + ax + b is divisible by (x 1) as well as (x 2).

The zeroes of (x – 1) and (x – 2) are 1, 2.

Let's substitute these values in x³ – 10x² + ax + b :

Case 1 x = 1 :

➺ x³ – 10x² + ax + b = 0

➺ ( 1 )³ – 10 ( 1 )² + a ( 1 ) + b = 0

➺ 1 – 10 + a + b = 0

➺ a + b – 9 = 0

➺ a + b = 9 . . . . . ( eq . 1)

Case 2 x = 2 :

➺ x³ – 10x² + ax + b = 0

➺ ( 2 )³ – 10 ( 2 )² + a ( 2 ) + b = 0

➺ 8 – 10 ( 4 ) + 2a + b = 0

➺ 8 – 40 + 2a + b = 0

➺ – 32 + 2a + b = 0

➺ 2a + b = 32 . . . . . (eq . 2)

Getting the value of 'a' from (1) :

➺ a + b = 9

a = 9 – b

Substituting this value in (2) :

➺ 2a + b = 32

➺ 2 (9 – b) + b = 32

➺ 18 – 2b + b = 32

➺ – b = 32 – 18

b = 14

Substituting this value in (1) to get 'a' :

➺ a + b = 9

➺ a + (– 14) = 9

➺ a – 14 = 9

➺ a = 9 + 14

a = 23

Therefore, the values of 'a' and 'b' are 23 and – 14.

Similar questions