x³+6x²+8x,x²-3x-10,x⁴-3x-10 Determine the sum of the three numbers....
Answers
Answer:
(i) x² - 5x + 6 and x² + 4x - 12
first of all factorize it ,
x² - 5x + 6 = (x - 2)(x - 3) -----(1)
x² + 4x - 12 = (x - 2)(x + 6) ------(2)
Now, LCM = GCD × prime factor of (1) × prime factor of (2)
Here, GCD = (x - 2)
prime factor of (1) = (x - 3)
prime factor of (2) = (x + 6)
Now, LCM = (x - 2)(x - 3)(x + 6)
(ii) x⁴ + 3x³ + 6x² + 5x + 3 and x⁴ + 2x² + x + 2 and GCD = x² + x + 1
factorise it ,
x⁴ + x³ + x² + 2x³ + 2x² + 2x + 3x² + 3x + 3
= x²(x² + x + 1) + 2x(x² + x + 1) + 3(x² + x + 1)
= (x² + x + 1)(x² + 2x + 3)
Similarly, x⁴ + 2x² + x + 2 = x⁴ + x³ + x² - x³ - x² - x + 2x² + 2x + 2
= x²(x² + x + 1) - x(x² + x + 1) + 2(x²+ x + 1)
= (x² + x + 1)(x² - x + 2)
Hence, LCM = (x² + x + 1)(x² + x + 2)(x² + 2x + 3)
(iii) 2x³ + 15x² + 2x - 35 = 2x³ + 14x² + x² + 7x - 5x - 35
= 2x²(x + 7) + x(x + 7) - 5(x + 7)
= (x + 7)(2x² + x - 5)
x³ + 8x² + 4x - 21 = x³ + 7x² + x² + 7x - 3x - 21
= x²(x + 7) + x(x + 7) - 3(x + 7)
= (x² + x - 3)(x + 7)
Hence, LCM = (x + 7)(x² + x - 3)(2x² + x - 5)
(iv) 2x³ - 3x² - 9x + 5 = 2x³ - x² - 2x² + x - 10x + 5
= x²(2x - 1) -x(2x - 1) -5(2x - 1)
= (2x - 1)(x² - x - 5)
2x⁴ - x³ - 10x² - 11x + 8 = 2x⁴ - x³ - 10x² + 5x - 16x + 8
= x³(2x - 1) -5x(2x - 1) -8(2x - 1)
= (2x - 1)(x³ - 5x - 8)
Hence, LCM = (2x - 1)(x³ - 5x - 8)(x² - x - 5)
thank u my dear lovely sister
