answer this question (class9)
no spamming ✖️
Answers
★ Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 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 a³ + b³ + c³ – 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 x³ – 3x² – 13x + 15 is exactly divisible by x² + 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 x³ – 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.