Question

what should be taken away from :5x³ - 6x²-3x+8 to get x⁴-1​

Answer 1

$- x {}^{4} + 5x {}^{3} - 6x {}^{2} - 3x + 8 \: \: \: \: \: \: \: \\ should \: be \: taken \: away$

Answer 2

(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)