Question

Factorise: x²+4abx-(a²-b²) ²​

Answer 1

Step-by-step explanation:

What is the factorisation of (a−2b)2−4a+8b?

A2A

(a−2b)2−4a+8b

=(a−2b)(a−2b)−4(a−2b)=(a−2b)(a−2b−4)

=(a−2b)(a−2(b+2))

Answer 2

Answer:

In order to factor this you are looking for:

(x−m)(x−n)=x2+4abx−(a2−b2)2

Where m and n are the roots of our equation.

For this, let's work with the quadratic formula. I'm going to do a partial substitution and see where it takes me:

−(4ab)±d−−√2⋅1

=−2ab±d√2

d being the determinant. Okay. What then is d?

d=(4ab)2–4(1)(−(a2−b2)2)

=16a2b2+4(a4–2a2b2+b4)

=4a4–8a2b2+16a2b2+4b4

=4a4+8a2b2+4b4

=4(a2+b2)2

And let's just figure out the ± portion of the equation:

d√2=4(a2+b2)2√2 v

=a2+b2

Which makes our roots:

−2ab+a2+b2=(a−b)2

−2ab−a2+b2

=−(a2+2ab+b2)=−(a+b)2

And that makes our factorization:

x2+4abx−(a2−b2)2

=(x−(a−b)2)(x+(a+b)2)

Step-by-step explanation:

hope it will help you

mark me as brainliest