Question

factorize completely the expression(h+2k)²+4k²-h²​

Answer 1

Answer:

$\rm 4k(2k+ h)$

Step-by-step explanation:

A expression is given to us and we need to factorise it . The given expression is ,

$\rm\implies (h+2k)^2+4k^2-h^2$

Open the brackets using the identity, (a +b)² = a² + b² + 2ab , we have ;

$\rm\implies h^2+4k^2+4hk + 4k^2 - h^2$

Simplify ,

$\rm\implies 8k^2+4hk$

In the above expression 4k is common in both the terms , so that ,

$\rm\implies \boxed{\pink{\frak{ 4k(2k+h) }}}$

Answer 2

The expression factorized completely is (h +2k)[(h+2k) + (2k-h)]

From the question,

We are to factorize the expression (h+2k)²+4k²-h²​ completely

The expression can be factorized as shown below

(h+2k)²+4k²-h²​ becomes

(h+2k)² + 2²k²-h²​

(h+2k)² + (2k)²-h²​

Using difference of two squares

The expression (2k)²-h²​ = (2k+h)(2k-h)

Then,

(h+2k)² + (2k)²-h²​ becomes

(h+2k)² + (2k +h)(2k-h)

This can be written as

(h+2k)² + (h +2k)(2k-h)

Now,

Factorizing, we get

(h +2k)[(h+2k) + (2k-h)]

Hence, the expression factorized completely is (h +2k)[(h+2k) + (2k-h)]

Learn more here:https://brainly.com/question/12486387