Louis calculates that the area of a rectangle is represented by the equation 3x4+7x2=2. Did is calculate right? Explain.
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.
Solve f(x)=3x4−x2−14. (#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.
Now, because neither factor can be factored further and there is no x−term, we can set each equal to zero and solve.
x2+2=0x2=−2andx=±−2−−−√ or ±i2√3x2−7=03x2=7x2=73x=±73−−√ or ±21−−√3
Including the imaginary solutions, there are four, which is what we would expect because the degree of this function is four.
Find all the solutions of the function g(x)=x4+21x2+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.
Now, set each factor equal to zero and solve.
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, 2x4−18x3+38x2+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.
Solving for x we get
Find all the solutions to the following functions.
3. Find the equation of a function with roots 4, 2√ and 1−i.
1. First, graph the function.
Using the Rational Root Theorem, the possible realistic zeros could be 125, 1 or 4. Let’s try these three possibilities using synthetic division.
Of these three possibilities, only 4 is a zero. The leftover polynomial, 25x2−20x+1 is not factorable, so we need to do the Quadratic Formula to find the last two zeros.
x=20±202−4(25)(1)−−−−−−−−−−−√2(25)=20±400−100−−−−−−−−√50=20±103√50 or 2±3√5≈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)=4x4+35x2−9 is factorable. ac=−36.
Setting each factor equal to zero, we have:
∗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, 2√,−2√,1+i,1−i. Multiply all 5 roots together.
Find all solutions to the following functions. Use any method.
Find a function with the following roots.
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 f(x)=x5+x3+8x2+8.