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Fundamental Theorem of Algebra

Practice Fundamental Theorem of Algebra
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Finding Imaginary Solutions

Louis calculates that the area of a rectangle is represented by the equation 3x^4 + 7x^2 = 2 . Did is calculate right? Explain.

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James Sousa: Ex 4: Find the Zeros of a Polynomial Function with Imaginary Zeros


In #12 from the previous problem set, there are two imaginary solutions. Imaginary solutions always come in pairs. To find the imaginary solutions to a function, use the Quadratic Formula. If you need a little review on imaginary numbers and how to solve a quadratic equation with complex solutions see the Quadratic Equations chapter.

Example A

Solve f(x)=3x^4-x^2-14 . (#12 from the previous problem set.)

Solution: First, this quartic function can be factored just like a quadratic equation. See the Factoring Polynomials in Quadratic Form concept from this chapter for review.

f(x) &= 3x^4-x^2-14\\0 &= 3x^4-7x^2+6x^2-14\\0 &= x^2(3x^2-7)+2(3x^2-7)\\0 &= (x^2+2)(3x^2-7)

Now, because neither factor can be factored further and there is no x- term, we can set each equal to zero and solve.

& && 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}

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

Example B

Find all the solutions of the function g(x)=x^4+21x^2+90 .

Solution: When graphed, this function does not touch the x- 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.

g(x) &= x^4+21x^2+90\\0 &= (x^2+6)(x^2+15)

Now, set each factor equal to zero and solve.

& 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}

Example C

Find the function that has the solution 3, -2, and 4 + i .

Solution: Notice that one of the given solutions is imaginary. Imaginary solutions always come in pairs, so 4 - i is also a factor, they are complex conjugates. Now, translate each solution into a factor and multiply them all together.

Any multiple of this function would also have these roots. For example, 2x^4-18x^3+38x^2+62x-204 would have these roots as well.

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

3x^4 + 7x^2 = 2\\= 3x^4 + 7x^2 - 2 = 0\\3x^4 + 7x^2 - 2 = (3x^2 + 1)(x^2 + 2) = 0

Solving for x we get

 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}

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

Guided Practice

Find all the solutions to the following functions.

1. f(x)=25x^3-120x^2+81x-4

2. f(x)=4x^4+35x^2-9

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


1. First, graph the function.

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

Of these three possibilities, only 4 is a zero. The leftover polynomial, 25x^2-20x+1 is not factorable, so we need to do the Quadratic Formula to find the last two zeros.

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

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

2. f(x)=4x^4+35x^2-9 is factorable. ac = -36 .

& 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)

Setting each factor equal to zero, we have:

& && 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}

^* This problem could have also been done by using the same method from #1.

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

& (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


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

  1. f(x)=x^4+x^3-12x^2-10x+20
  2. f(x)=4x^3-20x^2-3x+15
  3. f(x)=2x^4-7x^2-30
  4. f(x)=x^3+5x^2+12x+18
  5. f(x)=4x^4+4x^3-22x^2-8x+40
  6. f(x)=3x^4+4x^2-15
  7. f(x)=2x^3-6x^2+9x-27
  8. f(x)=6x^4-7x^3-280x^2-419x+280
  9. f(x)=9x^4+6x^3-28x^2+2x+11
  10. f(x)=2x^5-19x^4+30x^3+97x^2-20x+150

Find a function with the following roots.

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

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