Question #bced5

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Answer 1 · 2017-11-21T23:57:31

See a solution process below:

We can rewrite this function using this rule for quadratics:

$x^{2} - y^{2} = (x + y) (x - y)$ $color(red)(x)^2 - color(blue)(y)^2 = (color(red)(x) + color(blue)(y))(color(red)(x) - color(blue)(y))$

$f (x) = - 3 x^{4} (x^{2} - 9)$ $f(x) = -3x^4(color(red)(x)^2 - color(blue)(9))$

$f (x) = - 3 x^{4} (x^{2} - 3^{2})$ $f(x) = -3x^4(color(red)(x)^2 - color(blue)(3)^2)$

$f (x) = - 3 x^{4} (x + 3) (x - 3)$ $f(x) = -3x^4(color(red)(x) + color(blue)(3))(color(red)(x) - color(blue)(3))$

Now, we can solve each term on the left side of the function for $0$ $0$ to find each zero of the function:

Solution 1:

$- 3 x^{4} = 0$ $-3x^4 = 0$

$\frac{- 3 x^{4}}{- 3} = \frac{0}{- 3}$ $(-3x^4)/color(red)(-3) = 0/color(red)(-3)$

$(color(red)(cancel(color(black)(-3)))x^4)/cancel(color(red)(-3)) = 0$

$x^4 = 0$

$root(4)(x^4) = root(4)(0)$

$x = 0$

Solution 2:

$x + 3 = 0$

$x + 3 - color(red)(3) = 0 - color(red)(3)$

$x + 0 = -3$

$x = -3$

$x = 0$

Solution 3:

$x - 3 = 0$

$x - 3 + color(red)(3) = 0 + color(red)(3)$

$x - 0 = 3$

$x = 3$

The Solutions Are: $x = {-3, 0, 3}$

The graph of the function touch the $x$ -axis at $0$ :

graph{y + 3x^6 - 27x^4 = 0 [-10, 10, -50, 400]}