without actual decision prove that x⁴+2x³-2x²+2x-3 is exactly divisible by x²+2x-3
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Answer:
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.
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