So the greatest common factor 48 and 120 is 24.
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
Step-by-step explanation:
g(f(x))= 2(x + 1)
= 2x + 2
Answer:
a) (x + 5) (x - 5)
b) (x + 5i) (x - 5i)
c) (x + (5i/2)) (x - (5i/2))
d) (x-1)(x-1)
e) x +i√3 +1) (x -i√3+1)
Step-by-step explanation:
To solve this, we will need to factorize each quadratic function making it equal to zero first and then proceeding to find x
a) f(x) = x²-25
x²-25 = 0
⇒(x + 5) (x - 5)
b) f(x)=x²+25
x² + 25 = 0
x²= -25
x = √-25
x = √25i
x = ±5i
⇒(x + 5i) (x - 5i)
c) f(x)=4x²+25
4x²+25 = 0
4x²= -25
x² = -25/4
x = ±√(-25/4)
x = ±(√25i)/2
x = ±5i /2
⇒(x + (5i/2)) (x - (5i/2))
d) f(x)=x²-2x+1
x²-2x+1 = 0
⇒(x - 1)²
e) f(x)=x²-2x+4
x²-2x+4 = 0
x²-2x = -4
x²-2x +1 = -4 +1
x²-2x + 1 = -3
(x-1)² +3 = 0
(x-1)²= -3
x-1 = √-3
x = ±√3i +1
⇒(x +i√3 +1) (x -i√3+1)
