Louis calculates that the area of a rectangle is represented by the equation . Did is calculate right? Explain.
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 . (#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 term, we can set each equal to zero and solve.
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 .
Solution: When graphed, this function does not touch the 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 .
Solution: Notice that one of the given solutions is imaginary. Imaginary solutions always come in pairs, so 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, 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
All of the solutions are imaginary and the area of a rectangle must have real solutions. Therefore Louis did not calculate correctly.
Find all the solutions to the following functions.
3. Find the equation of a function with roots 4, and .
1. First, graph the function.
Using the Rational Root Theorem, the possible realistic zeros could be , 1 or 4. Let’s try these three possibilities using synthetic division.
Of these three possibilities, only 4 is a zero. The leftover polynomial, is not factorable, so we need to do the Quadratic Formula to find the last two zeros.
Helpful Hint: Always find the decimal values of each zero to make sure they match up with the graph.
2. is factorable. .
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, . 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 .