Question

Which of the following is not a factor of x4-4abc2-(a2-b2)2

Answer 1

You didn't give options in Ur question. Ur question is also wrong .

Question:

x⁴-4abx²-(a²-b²)²

Solution:


x⁴-4abx²-(a²-b²)²

Let x²= p

p²-4abp -(a²-b²)²=0

On comparing with ax²+ bx+c=0

a= 1, b= -4ab , c = -(a²-b²)²

P = -b ±(√b²-4ac)/2a
[ By quadratic Formula]

p= [-(-4ab) ±√ (-4ab)²- 4×1×-(a²-b²)²]/2×1

p= [4ab±√ (4ab)²+4(a²-b²)²]/2

p= [4ab± 2 √ (4a²b²)+(a²+ b² -2a²b²)]/2

p= [4ab± 2 √ 4a²b² -2a²b²+a²+ b²]/2

p= [4ab± 2 √ 2a²b²+a²+ b²]/2

p= [4ab± 2 √(a²+ b²+2a²b²)]/2

p = 4ab ±2√ (a²+b²)²/2

p = 4ab ±2 (a²+b²)/2

p = 2ab ±(a²+b²)

= 2ab +a²+b², 2ab -(a²+b²)

P= (a+b)², 2ab -a²-b²

P= (a+b)², -(a-b)²

p-(a+b)² , p+(a-b)²

x²-(a+b)² , x²+(a-b)²

(x)²-(a+b)² , x²+(a-b)²

(x-(a+b)), (x+(a+b)) x²+(a-b)²

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Hope this will help you...

Answer 2

Answer:-

Question:

x⁴-4abx²-(a²-b²)²

Solution:

x⁴-4abx²-(a²-b²)²

Let x²= p

p²-4abp -(a²-b²)²=0

On comparing with ax²+ bx+c=0

a= 1, b= -4ab , c = -(a²-b²)²

P = -b ±(√b²-4ac)/2a

[ By quadratic Formula]

p= [-(-4ab) ±√ (-4ab)²- 4×1×-(a²-b²)²]/2×1

p= [4ab±√ (4ab)²+4(a²-b²)²]/2

p= [4ab± 2 √ (4a²b²)+(a²+ b² -2a²b²)]/2

p= [4ab± 2 √ 4a²b² -2a²b²+a²+ b²]/2

p= [4ab± 2 √ 2a²b²+a²+ b²]/2

p= [4ab± 2 √(a²+ b²+2a²b²)]/2

p = 4ab ±2√ (a²+b²)²/2

p = 4ab ±2 (a²+b²)/2

p = 2ab ±(a²+b²)

= 2ab +a²+b², 2ab -(a²+b²)

P= (a+b)², 2ab -a²-b²

P= (a+b)², -(a-b)²

p-(a+b)² , p+(a-b)²

x²-(a+b)² , x²+(a-b)²

(x)²-(a+b)² , x²+(a-b)²

(x-(a+b)), (x+(a+b)) x²+(a-b)²

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Hope it's help You❤️