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zhenek [66]
3 years ago
11

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

Mathematics
2 answers:
slega [8]3 years ago
8 0

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) }}}

julsineya [31]3 years ago
5 0

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:brainly.com/question/12486387

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