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$$P(x)=16x^4-81$$

I know that this factors out as:

$$P(x)=16(x-\frac { 3 }{ 2 } )^4$$

What I don't understand is the four different zeros of the polynomial...I see one zero which is $\frac { 3 }{ 2 }$ but not the three others.

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  • $\begingroup$ Double check your factorization, its incorrect. $\endgroup$
    – Loocid
    Sep 23 '14 at 2:50
  • $\begingroup$ It's not fully factored you mean? $\endgroup$ Sep 23 '14 at 2:51
  • $\begingroup$ @Cherry_Developer No it should be: $P(x)=(2x-3)(2x+3)(4x^2+9)$ $\endgroup$
    – Rivasa
    Sep 23 '14 at 2:58
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    $\begingroup$ @Cherry_Developer I mean $16(x-\frac{3}{2})^4 = 81-216x+216 x^2-96x^3+16x^4$ not $16x^4-81$. That's why you are probably getting confused. Always double check your factorization where possible :). $\endgroup$
    – Loocid
    Sep 23 '14 at 2:59
  • $\begingroup$ Every one of us has made the mistake $(a+b)^2=a^2+b^2$, and we curse ourselves each time we do so. Now you have made the same mistake with fourth power instead of second. But don’t be doubly hard on yourself. $\endgroup$
    – Lubin
    Sep 23 '14 at 3:31
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Recognize $16x^4-81$ as a difference of two squares. Then factor into linear factors, $$ \begin{align*} P(x)&=16x^4-81\\ &=(4x^2)^2-9^2\\ &=(4x^2-9)(4x^2+9)\\ &=(2x-3)(2x+3)(2x+3i)(2x-3i). \end{align*} $$ Now $P(x)=0$ iff $$ 2x\pm3=0 \qquad\text{or}\qquad 2x\pm3i=0 $$ iff $$ x=\pm\frac32 \qquad\text{or}\qquad x=\pm\frac32i. $$ So we've found all four roots.

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Try making the substitutions $a=2x$ and $b=3$. Then we have that $$ 16x^4-81=a^4-b^4. $$ Recalling that $x^2-y^2=(x+y)(x-y)$, we have $$ a^4-b^4=[a^2-b^2](a^2+b^2)=[(a-b)(a+b)](a^2+b^2) $$ Plugging $2x$ in for $a$ and $3$ in for $b$ we have $$ (a-b)(a+b)(a^2+b^2)=(2x-3)(2x+3)((2x)^2+3^2)=(2x-3)(2x+3)(4x^2+9). $$ If you set each of these three factors equal to zero you will find the four roots you are looking for.

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  • $\begingroup$ $4x^2+9$ becomes: $x^{ 2 }=\frac { -9 }{ 4 }$ How does that give me the right answer?? $\endgroup$ Sep 23 '14 at 3:15
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    $\begingroup$ From here we have $x=\pm\sqrt{-\frac{9}{4}}=\pm i\frac{3}{2}$. Alternatively, we can write $4x^2+9$ as $(2x)^2-(3i)^2=(2x-3i)(2x+3i)$. $\endgroup$
    – Eric
    Sep 23 '14 at 3:20
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You have to solve the expression $16x^4 - 81 = 0$, and you will get

$$ 16x^4 - 81 = (4x^2 - 9)(4x^2 + 9) = 0 $$

then you will find the other three roots you didn't find.

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The four zeros are $\frac{3}{2},\frac{-3}{2},\frac{3i}{2},\frac{-3i}{2}$

The $n$th root of a number has $n$ answers.

To find each value of the $n$th root of $x$, draw a unit circle. The principal root is simply horizontal. Take that line and rotate it about the origin by increments of $\dfrac{2\pi}{n}$ to get each value. The y axis is imaginary and the x axis is real.

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  • $\begingroup$ I know that. I just don't understand how to arrive at those four zeros. $\endgroup$ Sep 23 '14 at 2:51

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