QUESTION 16
The given function is,
[tex]y = 3x {(x + 2)}^{3} [/tex]
To find the zeroes of this function, we equate the function to zero to get,
[tex]3x {(x + 2)}^{3} = 0[/tex]
We now apply the zero product principle to get,
[tex]3x = 0 \: or \: {(x + 2)}^{3} = 0[/tex]
The second factor is repeating 3 times, therefore the root has a multiplicity of 3.
This implies that,
[tex]x = 0 \: or \: x = - 2[/tex]
-2 has a multiplicity of 3.
QUESTION 17.
The given expression is
[tex] {x}^{4} - 8 {x}^{2} + 16 = 0[/tex]
[tex] ({x}^{2}) ^{2} - 8 {x}^{2} + 16 = 0[/tex]
This is now a quadratic equation in x².
We split the middle term to get,
[tex] ({x}^{2}) ^{2} - 4 {x}^{2} - 4 {x}^{2} + 16 = 0[/tex]
We now factor to obtain,
[tex] {x}^{2} ( {x}^{2} - 4) - 4( {x}^{2} - 4) = 0[/tex]
This implies that,
[tex]( {x}^{2} - 4)( {x}^{2} - 4) = 0[/tex]
[tex]( {x}^{2} - 2^2)( {x}^{2} - 2^2) = 0[/tex]
We apply difference of two squares here to get,
[tex](x - 2)(x + 2)(x - 2)(x + 2) = 0[/tex]
This is the same as,
[tex](x - 2) ^{2} (x + 2) ^{2} = 0[/tex]
This time both roots have a multiplicity of 2.
Applying the zero product property, we obtain,
[tex](x - 2) ^{2} = 0 \: or \: (x + 2) ^{2} = 0[/tex]
This implies that,
[tex]x = 2 \: or \: - 2[/tex]
with a multiplicity of 2 each.
