Question

without actual division , prove that x⁴+2x³-2x²+2x-3 is exactly divisible by x²+2x-3 .​

Answer 1

Step-by-step explanation:

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,

Answer 2

Answer:

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