# Not understanding how to factor a polynomial completely

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

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.

• Double check your factorization, its incorrect. Commented Sep 23, 2014 at 2:50
• It's not fully factored you mean? Commented Sep 23, 2014 at 2:51
• @Cherry_Developer No it should be: $P(x)=(2x-3)(2x+3)(4x^2+9)$ Commented Sep 23, 2014 at 2:58
• @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 :). Commented Sep 23, 2014 at 2:59
• 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. Commented Sep 23, 2014 at 3:31

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.

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.

• $4x^2+9$ becomes: $x^{ 2 }=\frac { -9 }{ 4 }$ How does that give me the right answer?? Commented Sep 23, 2014 at 3:15
• From here we have $x=\pm\sqrt{-\frac{9}{4}}=\pm i\frac{3}{2}$. Alternatively, we can write $4x^2+9$ as $(2x)^2-(3i)^2=(2x-3i)(2x+3i)$.
– Eric
Commented Sep 23, 2014 at 3:20

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.

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.

• I know that. I just don't understand how to arrive at those four zeros. Commented Sep 23, 2014 at 2:51