Question

Find the zeros of each function. State the multiplicity of any multiple zeros. (Please show work)

Answer 1

QUESTION 16

The given function is,

$y = 3x {(x + 2)}^{3}$

To find the zeroes of this function, we equate the function to zero to get,

$3x {(x + 2)}^{3} = 0$

We now apply the zero product principle to get,

$3x = 0 \: or \: {(x + 2)}^{3} = 0$

The second factor is repeating 3 times, therefore the root has a multiplicity of 3.

This implies that,

$x = 0 \: or \: x = - 2$
-2 has a multiplicity of 3.

QUESTION 17.

The given expression is

${x}^{4} - 8 {x}^{2} + 16 = 0$

$({x}^{2}) ^{2} - 8 {x}^{2} + 16 = 0$

This is now a quadratic equation in x².

We split the middle term to get,

$({x}^{2}) ^{2} - 4 {x}^{2} - 4 {x}^{2} + 16 = 0$

We now factor to obtain,

${x}^{2} ( {x}^{2} - 4) - 4( {x}^{2} - 4) = 0$

This implies that,

$( {x}^{2} - 4)( {x}^{2} - 4) = 0$

$( {x}^{2} - 2^2)( {x}^{2} - 2^2) = 0$

We apply difference of two squares here to get,

$(x - 2)(x + 2)(x - 2)(x + 2) = 0$

This is the same as,

$(x - 2) ^{2} (x + 2) ^{2} = 0$

This time both roots have a multiplicity of 2.

Applying the zero product property, we obtain,

$(x - 2) ^{2} = 0 \: or \: (x + 2) ^{2} = 0$

This implies that,

$x = 2 \: or \: - 2$

with a multiplicity of 2 each.