What is expand (3(k+1)(k+2))/2 ?

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

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expand (3(k+1)(k+2))/2

expand

\frac{3 ( k + 1 ) ( k + 2 )}{2}

\frac{3 k ^{2}}{2} + \frac{9 k}{2} + 3

Solution steps

\frac{3 ( k + 1 ) ( k + 2 )}{2}

Expand

3 (k + 1) (k + 2) : 3 k^{2} + 9 k + 6

= \frac{3 k ^{2} + 9 k + 6}{2}

Apply the fraction rule:

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

\frac{3 k ^{2} + 9 k + 6}{2} = \frac{3 k ^{2}}{2} + \frac{9 k}{2} + \frac{6}{2}

= \frac{3 k ^{2}}{2} + \frac{9 k}{2} + \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= \frac{3 k ^{2}}{2} + \frac{9 k}{2} + 3

(f \circ g) (x - y)

16 (4 π - 3)

- 2 (2 x y + \frac{d}{d x} ((y) x^{2}))

(2 x + 3) \circ (x^{2} - 2 x + 4)

\frac{( - x + 20 ) ( x - 10 )}{2}

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