Question

Without actual division, prove that x4 + 2x3 – 2x2 + 2x – 3 is exactly divisible by

x

2

+ 2x – 3.

Answer 1

x² + 2x - 3
=> x² + 3x - x - 3
=> x(x + 3) - (x + 3)
=> (x + 3)(x - 1)


Now, by remainder theorem, remainder = 0

Taking, x + 3 as the factor

So, x = -3



x⁴ + 2x³ - 2x² + 2x - 3 = 0

(-3)⁴ + 2(3)³ - 2(-3)² + 2(-3) - 3 = 0

81 - 54 - 18 - 6 - 3 = 0

81 - 54 - 27 = 0

81 - 81 = 0

0 = 0


Hence, (x + 3) is the factor of given equation,



Now, checking for (x - 1) as a factor,
So, x = 1




x⁴ + 2x³ - 2x² + 2x - 3 = 0

(1)⁴ + 2(1)³ - 2(1)² + 2(1) - 3 = 0

1 + 2 - 2 + 2 - 3 = 0

3 - 3 = 0

0 = 0


Then, x - 1 is Also a factor,


As both (x - 1) and (x + 3) are factors, we can say that the given equation is divisible by x² +2x - 3


Hence, proved.






I hope this will help you


(-:

Answer 2

x² + 2x -3 = 0
= x² + 3x - x -3 = 0
= x(x +3) -1(x + 3)= 0
= (x - 1)(x + 3) = 0x = 1; x = -3

put the values of x in the equation x⁴ + 2x³ - 2x² + 2x -3  if the equation is equal to zero for any values of x (1 and/or -3 )then we can say that its is exactly divisible by the eqn x² + 2x - 3. 
When x =1: 1⁴ + 2×1³ - 2×1² + 2×1 -3 = 1+ 2 - 2 +2 -3 = 0
When x = -3:
(-3)⁴ + 2×(-3)³ - 2×(-3)² + 2×(-3) -3
= 81 - 54 - 18 - 6 -3
= 0
Therefore x⁴ + 2x³ - 2x² + 2x -3  is completely divisible by x² + 2x - 3 for x =1 and x= -3;