What is (64-48x^2+12x^4-x^6)*(-1/3) ?

The solution to (64-48x^2+12x^4-x^6)*(-1/3) is -64/3+16x^2-4x^4+(x^6)/3

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simplify (64-48x^2+12x^4-x^6)*(-1/3)

simplify

(64 - 48 x^{2} + 12 x^{4} - x^{6}) \cdot (- \frac{1}{3})

- \frac{64}{3} + 16 x^{2} - 4 x^{4} + \frac{x ^{6}}{3}

Solution steps

(64 - 48 x^{2} + 12 x^{4} - x^{6}) (- \frac{1}{3})

Apply rule:

a (- b) = - ab

(64 - 48 x^{2} + 12 x^{4} - x^{6}) (- \frac{1}{3}) = - (64 - 48 x^{2} + 12 x^{4} - x^{6}) \frac{1}{3}

= - (64 - 48 x^{2} + 12 x^{4} - x^{6}) \frac{1}{3}

= - \frac{1}{3} (64 - 48 x^{2} + 12 x^{4} - x^{6})

Expand

- \frac{1}{3} (64 - 48 x^{2} + 12 x^{4} - x^{6}) : - \frac{64}{3} + 16 x^{2} - 4 x^{4} + \frac{x ^{6}}{3}

= - \frac{64}{3} + 16 x^{2} - 4 x^{4} + \frac{x ^{6}}{3}

H (t - a) (t - a)

H (t - a) (t - b)

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

(x^{2} + y^{2}) (y, x, e^{z}) + (x y + e^{z}) (2 x, 2 y, 0)

arccos (0)

What is (64-48x^2+12x^4-x^6)*(-1/3) ?
The solution to (64-48x^2+12x^4-x^6)*(-1/3) is -64/3+16x^2-4x^4+(x^6)/3