What is expand 2((1-3y)/2)^2 ?

The solution to expand 2((1-3y)/2)^2 is (9y^2)/2-3y+1/2

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expand 2((1-3y)/2)^2

expand

2 (\frac{1 - 3 y}{2})^{2}

\frac{9 y ^{2}}{2} - 3 y + \frac{1}{2}

Solution steps

2 (\frac{1 - 3 y}{2})^{2}

(\frac{1 - 3 y}{2})^{2} = \frac{1 - 6 y + 9 y ^{2}}{2 ^{2}}

= 2 \cdot \frac{9 y ^{2} - 6 y + 1}{2 ^{2}}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 1 - 6 y + 9 y ^{2} ) \cdot 2}{2 ^{2}}

Cancel the common factor:

2

= \frac{1 - 6 y + 9 y ^{2}}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 - 6 y + 9 y ^{2}}{2} = \frac{1}{2} - \frac{6 y}{2} + \frac{9 y ^{2}}{2}

= \frac{1}{2} - \frac{6 y}{2} + \frac{9 y ^{2}}{2}

Divide the numbers:

\frac{6}{2} = 3

= \frac{1}{2} - 3 y + \frac{9 y ^{2}}{2}

Rewrite in standard form

= \frac{9 y ^{2}}{2} - 3 y + \frac{1}{2}

\frac{2 π}{365} (90 - 101)

\frac{a \cdot b}{a ( 1 + c ) + d}

x^{x} ln (x) (ln (x) + 1) + x^{x - 1}

8 cos (2 (x + 1))

\frac{3 ( x - 5 )}{8}

What is expand 2((1-3y)/2)^2 ?
The solution to expand 2((1-3y)/2)^2 is (9y^2)/2-3y+1/2