Louis calculates that the area of a rectangle is represented by the equation \begin{align*}3x^4 + 7x^2 = 2\end{align*}. Did Louis calculate it right? Explain based on the degree and zeros of the function.

### Imaginary Solutions

Remember, imaginary solutions always come in pairs. To find the imaginary solutions to a function, use the Quadratic Formula.

Let's solve \begin{align*}f(x)=3x^4-x^2-14\end{align*}.

First, this quartic function can be factored just like a quadratic equation.

\begin{align*}g(x)=x^4+21x^2+90\end{align*}

Now, because neither factor can be factored further and there is no \begin{align*}x-\end{align*}term, we can set each equal to zero and solve.

\begin{align*}& && 3x^2-7=0\\ & x^2+2=0 && 3x^2=7\\ & x^2=-2 \qquad \qquad \qquad \quad and && x^2=\frac{7}{3}\\ & x=\pm \sqrt{-2} \ or \ \pm i \sqrt{2} && x=\pm \sqrt{\frac{7}{3}} \ or \ \pm \frac{\sqrt{21}}{3}\end{align*}

Including the imaginary solutions, there are four, which is what we would expect because the degree of this function is four.

Now, let's find all the solutions of the function \begin{align*}g(x)=x^4+21x^2+90\end{align*}.

When graphed, this function does not touch the \begin{align*}x-\end{align*}axis. Therefore, all the solutions are imaginary. To solve, this function can be factored like a quadratic equation. The factors of 90 that add up to 21 are 6 and 15.

\begin{align}g(x) &= x^4+21x^2+90\\ 0 &= (x^2+6)(x^2+15)\end{align}

Now, set each factor equal to zero and solve.

\begin{align*}& x^2+6=0 && x^2+15=0\\ & x^2=-6 \qquad \qquad \qquad and && x^2=-15\\ & x=\pm i \sqrt{6} && x=\pm i \sqrt{15}\end{align*}

Finally, let's find the function that has the solution 3, -2, and \begin{align*}4 + i\end{align*}.

Notice that one of the given solutions involves an imaginary number. Imaginary and complex solutions always come in pairs, so \begin{align*}4 - i\end{align*} is also a factor. The two factors are **complex conjugates.** Translate each solution into a factor and multiply them all together.

Any multiple of this function would also have these roots. For example, \begin{align*}2x^4-18x^3+38x^2+62x-204\end{align*} would have these roots as well.

### Examples

#### Example 1

Earlier, you were asked to determine if Louis calculated his work correctly.

First we need to change the equation to standard form. Then we can factor it.

\begin{align}3x^4 + 7x^2 &= 2\\ 3x^4 + 7x^2 - 2 &= 0\\ (3x^2 + 1)(x^2 + 2) &= 0 \end{align}

Solving for *x* we get

\begin{align*} 3x^2+1=0 && x^2+2=0\\ & x^2=\frac{-1}{3} \qquad \qquad \qquad and && x^2=-2\\ & x=\pm i \sqrt{\frac{1}{3}} && x=\pm i \sqrt{2}\end{align*}

All of the solutions are imaginary and the area of a rectangle must have real solutions. Therefore Louis did not calculate correctly.

#### Example 2

Find all the solutions to the following function: \begin{align*}f(x)=25x^3-120x^2+81x-4\end{align*}.

First, graph the function.

Using the Rational Root Theorem, the possible realistic zeros could be \begin{align*}\frac{1}{25}\end{align*}, 1, or 4. Let’s try these three possibilities using synthetic division.

Of these three possibilities, only 4 is a zero. The leftover polynomial, \begin{align*}25x^2-20x+1\end{align*} is not factorable, so we need to use the Quadratic Formula to find the last two zeros.

\begin{align}x &= \frac{20 \pm \sqrt{20^2-4(25)(1)}}{2(25)}\\ &= \frac{20 \pm \sqrt{400-100}}{50}\\ &=\frac{20 \pm 10 \sqrt{3}}{50} \ or \ \frac{2 \pm \sqrt{3}}{5} \approx 0.746 \ and \ 0.054 \end{align}

**Helpful Hint:** Always find the decimal values of each zero to make sure they match up with the graph.

#### Example 3

Find all the solutions to the following function: \begin{align*}f(x)=4x^4+35x^2-9\end{align*}.

\begin{align*}f(x)=4x^4+35x^2-9\end{align*} is factorable. \begin{align*}ac = -36\end{align*}.

\begin{align*} 4x^4+35x^2-9\\ 4x^4+36x^2-x^2-9\\ 4x^2(x^2+9)-1(x^2+9)\\ (x^2+9)(4x^2-1)\end{align*}

Setting each factor equal to zero, we have:

\begin{align*}& && 4x^2-1=0\\ & x^2+9=0 && 4x^2=1\\ & x^2=-9 \quad \qquad \qquad or && x^2=\frac{1}{4}\\ & x=\pm 3i && x=\pm \frac{1}{2}\end{align*}

#### Example 4

Find the equation of a function with roots 4, \begin{align*}\sqrt{2}\end{align*} and \begin{align*}1-i\end{align*}.

Recall that irrational and imaginary roots come in pairs. Therefore, all the roots are 4, \begin{align*}\sqrt{2}, {\color{red}-\sqrt{2}},1+i,{\color{red}1-i}\end{align*}. Multiply all 5 roots together.

\begin{align*} (x-4)(x-\sqrt{2})(x+\sqrt{2})(x-(1+i))(x-(1-i))\\ (x-4)(x^2-2)(x^2-2x+2)\\ (x^3-4x^2-2x+8)(x^2-2x+2)\\ x^5-6x^4+8x^3-4x^2-20x+16\end{align*}

### Review

Find all solutions to the following functions. Use any method.

- \begin{align*}f(x)=x^4+x^3-12x^2-10x+20\end{align*}
- \begin{align*}f(x)=4x^3-20x^2-3x+15\end{align*}
- \begin{align*}f(x)=2x^4-7x^2-30\end{align*}
- \begin{align*}f(x)=x^3+5x^2+12x+18\end{align*}
- \begin{align*}f(x)=4x^4+4x^3-22x^2-8x+40\end{align*}
- \begin{align*}f(x)=3x^4+4x^2-15\end{align*}
- \begin{align*}f(x)=2x^3-6x^2+9x-27\end{align*}
- \begin{align*}f(x)=6x^4-7x^3-280x^2-419x+280\end{align*}
- \begin{align*}f(x)=9x^4+6x^3-28x^2+2x+11\end{align*}
- \begin{align*}f(x)=2x^5-19x^4+30x^3+97x^2-20x+150\end{align*}

Find a function with the following roots.

- \begin{align*}4, i\end{align*}
- \begin{align*}-3, -2i\end{align*}
- \begin{align*}\sqrt{5}, -1 + i\end{align*}
- \begin{align*}2, \frac{1}{3}, 4-\sqrt{2}\end{align*}
**Writing:**Write down the steps you use to find all the zeros of a polynomial function.**Writing:**Why do imaginary and irrational roots always come in pairs?**Challenge:**Find all the solutions to \begin{align*}f(x)=x^5+x^3+8x^2+8\end{align*}.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 6.12.