What is factor 4y^2+4yz+z^2-1 ?

The solution to factor 4y^2+4yz+z^2-1 is (2y+z+1)(2y+z-1)

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factor 4y^2+4yz+z^2-1

factor

4 y^{2} + 4 yz + z^{2} - 1

(2 y + z + 1) (2 y + z - 1)

Solution steps

4 y^{2} + 4 yz + z^{2} - 1

Factor

4 y^{2} + 4 yz + z^{2} : (2 y + z)^{2}

= (2 y + z)^{2} - 1

Rewrite

1

1^{2}

= (2 y + z)^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(2 y + z)^{2} - 1^{2} = ((2 y + z) + 1) ((2 y + z) - 1)

= ((2 y + z) + 1) ((2 y + z) - 1)

Remove parentheses:

(a) = a

= (2 y + z + 1) (2 y + z - 1)

0 = - \frac{3}{50} (x - 20)^{2} + 24

- 5 \frac{- x + 4}{5} > 11 - 3 x

50 \cdot (1 - e^{\frac{- 4}{8 \cdot 6}})

8 - 50 x^{4}

\frac{2 x ^{3} y ^{5}}{5 ( x ^{4} y ^{2} ) ^{3}}

What is factor 4y^2+4yz+z^2-1 ?
The solution to factor 4y^2+4yz+z^2-1 is (2y+z+1)(2y+z-1)