Kevin B. answered • 10/22/22
A Specialist in Math and Physics
I've added plenty of notes to explain some of the things I've seen people get confused about. I've also provided very detailed simplification steps. Please skip any of these notes as needed.
The given value
n = 3
means that we need to determine a 3rd-degree polynomial (n: degree-- the highest power of the independent variable x in the function) which satisfies the remaining given values.
The polynomial will have the form
f(x) = ax3 + bx2 + cx + d
or
f(x) = (x - x1)(x - x2)(x - x3)
where x1, x2, x3 are the zeroes of the function. We remember that the number of zeroes in a polynomial function is the same as the degree, which is 3 in this problem. The goal of this problem is to find the values for these zeroes and plug them into the above function. In this case, the function itself is the answer to the problem.
SPECIAL NOTE: Even if there are repeated zeroes (this is called "multiplicity"), we can assume that the function will take the above form.
We have been given two of the zeroes, so we can plug them in directly:
x1 = -4
x2 = 8+2i
NOTE: x2 is a complex zero, because it contains an imaginary term (an i term). Don't be intimidated, though-- we can still apply math and algebra rules to it just like any other number.
Put the zeroes into the function we are building:
f(x) = (x + 4)(x - [8+2i])(x - x3)
NOTE: When [ ] are used in an expression, it means the same thing as ( ). Using both can make the expression easier to understand visually.
Just from this one step, we are almost there. But we still don't know the value of x3, meaning we need one more piece of data to finish the problem. The key is the final given value:
f(-1) = 255
In other words, we are being told ahead of time that if we plug in a -1 for x into our mystery function, it will evaluate to 255. If we do this, the only remaining variable will be x3, and we can solve for it using basic algebra.
Here we go:
f(x) = (x + 4)(x - [8+2i])(x - x3)
so
f(-1) = (-1 + 4)(-1 - [8+2i])(-1 - x3)
= (3)(-9-2i)(-1 - x3) --addition
= (27+6i)(1 + x3) --distribute the 3 (negatives cancel)
=(27+6i) + (27+6i)x3 --distribute the (27+6i)
but since f(-1) = 255...
(27+6i) + (27+6i)x3 = 255
Look! An equation with only 1 variable means we can solve for the value of that variable.
NOTE: If you are confused about the previous step, remember the transitive property of equality: "If a = b and b = c, then a = c." In this case, if f(-1) = 255 and f(-1) = (a function), then (a function) = 255
To solve for x3 (remember to apply order of operations in reverse):
(27+6i) + (27+6i)x3 = 255
Subtract from both sides (27+6i)
(27+6i)x3 = 255 - (27+6i)
= 228 - 6i
Divide both sides by (27+6i)
x3 = (228 - 6i) / (27+6i)
This is our final zero. But imaginary numbers in denominators are messy and difficult to worth with, so we can change it to a slightly different form to fix this problem (you will generally not be given full credit if your answers have complex denominators):
x3 = (228 - 6i) / (27+6i)
= [ (228 - 6i) / (27+6i) ] * [ (27-6i) / (27-6i) ] --multiply the numerator and the denominator by (27+6i)
Why do we do this?
Any complex number multiplied by its complex conjugate (just flip the sign of the i term to get this) is a real (non-imaginary) number. If you are unclear on this, you will see why in a moment.
Why are we able to do this?
We can multiply the denominator by anything we want, as long as we multiply the numerator by the same thing. This is because multiplying values into the numerator and denominator of a fraction is the same thing as multiplying it by another fraction. Any fraction of the form A / A just equals 1. Therefore, multiplying the same number into the numerator and denominator does not change the value, just the representation of the value.
Continuing from above:
x3 = [ (228 - 6i) / (27+6i) ] * [ (27-6i) / (27-6i) ]
= [ (228 - 6i)*(27-6i) ] / [ (27+6i)*(27-6i) ]
= (6156 - 1368i - 162i - 36) / (729 + 162i - 162i + 36) --two FOILs/double-distributes
Remember that the definition of i is the square root of -1. Therefore, i * i is -1. Keep this in mind when carrying out these simplifications.
Continuing:
x3 = (6156 - 1368i - 162i - 36) / (729 + 162i - 162i + 36) --notice how the i term disappears from the denominator
= (6120 - 1530i) / 765 --addition
= 8 - 2i --division
We now have our final zero in a clean format. Add it to the function to get the answer:
f(x) = (x + 4)(x - [8+2i])(x - [8-2i])
or, to get it in its expanded form:
f(x) = (x + 4)(x - [8+2i])(x - [8-2i])
= (x2 - [8+2i]x + 4x - [32+8i])(x - [8-2i]) --FOIL/Double-distribute
= (x2 - [4+2i]x - [32+8i])(x - [8-2i]) --addition
= x3 - (4+2i)x2 - (32+8i)x - (8-2i)x2 + (4+2i)(8-2i)x + (32+8i)(8-2i) --more distribution
= x3 - (4+2i)x2 - (32+8i)x - (8-2i)x2 + (32-8i+16i+4)x + (256-64i+64i+16) --two more FOILs/double-distributes
= x3 - 12x2 + 4x + 272 --a bunch of addition
Now we have our function in two forms, both of which are often acceptable answers, but it's important to know if the instructor generally prefers one over the other.
To check our answer, we can just plug in the given values and see if we get the expected outputs. We know that
f(-1) = 255
So then
f(-1) = (-1)3 - 12(-1)2 + 4(-1) + 272
= -1 -12 -4 + 272 --apply exponents and multiply
= 255 --addition
Exactly as expected. We can also check the zeroes. Remember that a polynomial's "zeroes" are just the values of x where the function evaluates to 0. The problem specifies that we only have to verify the real zeroes, of which there is only one, x = -4:
f(-4) = 0
f(-4) = (-4)3 - 12(-4)2 + 4(-4) + 272
= -64 - 192 - 16 + 272 --apply exponents and multiply
= 0 --addition
so we know we did everything right.
SPECIAL NOTE: There is a way to save ourselves a lot of work by doing some extra thinking. At the beginning of the problem, we know the following:
f(x) = (x + 4)(x - [8+2i])(x - x3)
f(-1) = 255
We realize this means that there is no i term left after the function is evaluated. The only way for this to be true is if the last zero is the complex conjugate of the second zero (the two will be multiplied together, eliminating the i term). Therefore x3 must be 8-2i, and we can skip straight to verifying our answer. The instructor might not allow this shortcut, though.
----------------------------------------------------------------------------
One more example.
Find the polynomial that satisfies all of the following:
n = 3
Zeroes: 3, 1+2i
f(2) = -5
A 3rd-degree polynomial has the form
f(x) = (x - x1)(x - x2)(x - x3)
After substituting the given zeroes, we have
f(x) = (x - 3)(x - [1+2i])(x - x3)
Using f(2) = -5:
f(2) = (2 - 3)(2 - [1+2i])(2 - x3) = -5
(-1)(1-2i)(2 - x3) = -5
(-1+2i)(2 - x3) = -5
(-2+4i) + (1-2i)x3 = -5
(1-2i)x3 = -5 - (-2+4i)
x3 = (-3-4i) / (1-2i)
= ( [-3-4i][1+2i] ) / ( [1-2i][1+2i] )
= 5-10i / 5
= 1-2i
f(x) = (x - 3)(x - [1+2i])(x - [1-2i])
= (x - 3)(x2 - 2x + 5)
= x3 - 2x2 - 3x2 + 5x + 6x - 15
= x3 - 5x2 + 11x - 15
Now to verify:
f(2) = (2)3 - 5(2)2 + 11(2) - 15
= 8 - 20 + 22 - 15
= -5 --Good!
f(3) = (3)3 - 5(3)2 + 11(3) - 15
= 27 - 45 + 33 - 15
= 0 --Good!
All done.
