What is expand ((x^2+2x-6)^2)/4 ?

The solution to expand ((x^2+2x-6)^2)/4 is (x^4)/4+x^3-2x^2-6x+9

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expand ((x^2+2x-6)^2)/4

expand

\frac{( x ^{2} + 2 x - 6 ) ^{2}}{4}

\frac{x ^{4}}{4} + x^{3} - 2 x^{2} - 6 x + 9

Solution steps

\frac{( x ^{2} + 2 x - 6 ) ^{2}}{4}

(x^{2} + 2 x - 6)^{2} = x^{4} + 4 x^{3} - 8 x^{2} - 24 x + 36

= \frac{x ^{4} + 4 x ^{3} - 8 x ^{2} - 24 x + 36}{4}

Apply the fraction rule:

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

\frac{x ^{4} + 4 x ^{3} - 8 x ^{2} - 24 x + 36}{4} = \frac{x ^{4}}{4} + \frac{4 x ^{3}}{4} - \frac{8 x ^{2}}{4} - \frac{24 x}{4} + \frac{36}{4}

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{4} - \frac{8 x ^{2}}{4} - \frac{24 x}{4} + \frac{36}{4}

\frac{4 x ^{3}}{4} = x^{3}

\frac{8 x ^{2}}{4} = 2 x^{2}

\frac{24 x}{4} = 6 x

\frac{36}{4} = 9

= \frac{x ^{4}}{4} + x^{3} - 2 x^{2} - 6 x + 9

P (- 5, 4, 2) e Q (3, - 3, 9)

\frac{( 2 x - 1 ) ^{2}}{2 ( x - 1 )}

4 cos (2 (x - \frac{π}{3})) + 1

(1 - u - v, 1 - u, u) \cdot (1, 1, 1)

(r^{2} - 4 r - 5) \cdot e^{r x}

What is expand ((x^2+2x-6)^2)/4 ?
The solution to expand ((x^2+2x-6)^2)/4 is (x^4)/4+x^3-2x^2-6x+9