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

The solution to expand (2x-3λ(x-1)^2)/((2λ+2)y) is (2x-3λx^2+6λx-3λ)/(2λy+2y)

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expand (2x-3λ(x-1)^2)/((2λ+2)y)

expand

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

\frac{2 x - 3 λ x ^{2} + 6 λ x - 3 λ}{2 λ y + 2 y}

Solution steps

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

(x - 1)^{2} = x^{2} - 2 x + 1

= \frac{2 x - 3 λ ( x ^{2} - 2 x + 1 )}{y ( 2 λ + 2 )}

Expand

(2 λ + 2) y : 2 λ y + 2 y

= \frac{2 x - 3 λ ( x ^{2} - 2 x + 1 )}{2 λ y + 2 y}

Expand

2 x - 3 λ (x^{2} - 2 x + 1) : 2 x - 3 λ x^{2} + 6 λ x - 3 λ

= \frac{2 x - 3 λ x ^{2} + 6 λ x - 3 λ}{2 λ y + 2 y}

\frac{1 + j}{2 ( 2 - 2 j )}

(7.5, - 2.2) y (4.8, - 5.6)

\frac{2 s + 9}{( s + 4 ) ^{2}}

\frac{2 ( 50 \cdot 5 ^{\frac{1}{2}} - 18 \cdot 3 ^{\frac{1}{2}} )}{15}

\frac{10 - 7 x}{( x + 2 ) ( x - 2 )}

What is expand (2x-3λ(x-1)^2)/((2λ+2)y) ?
The solution to expand (2x-3λ(x-1)^2)/((2λ+2)y) is (2x-3λx^2+6λx-3λ)/(2λy+2y)