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)]
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Answer:
3,2 and -2
because you can't make denominator zero
and only these will make deno. zero
M equals 3x+21 because you add 2x+1x=3x.then you add 14+7=21.then you combine them together to get 3x+21
3(1) - 8 = 3 - 8 = -5
just plug in 1 for x that's what f(1) means
Answer :
inequality form: k ≥ -4
OR
interval form: [-4, ∞)
