Get the answer to your homework problem.
Try Numerade free for 7 days
Michigan State University
We don’t have your requested question, but here is a suggested video that might help.
Related Question
Write a polynomial function of least degree with integral coefficients the zeros of which include 2 and 4$i .$
Discussion
You must be signed in to discuss.
Video Transcript
So we want to find a polynomial that has zeros at two and four I and we want them to have integer coefficients. So we actually know that if we have a complex zero that they have to come as conjugate pairs, complex conjugate pairs. So we know that this also must be a zero. So we know our with the least degree is a third degree polynomial and there are infinitely many possibilities with a multiplier. But why don't we use this? Let's have X minus this zero an x minus This zero next minus a negative. For I guess is that an X minus that zero. Now that polynomial will work. We could also put a coefficient here in front like a two or a negative two of five or seven etcetera to make the energy, er the polynomial expand or contract up and down. But now we need to put it in polynomial form. So let's multiply this out. And when we multiply these two together, first we end up with X squared and we get four i X and negative for I excel those cancel out and then a negative four i and a positive four I When we get negative 16 I squared. That becomes plus 16. So the products of these to the eyes cancel out. And so now we just need to multiply these two together, and then we'll have our answer. So we have X to the third. Oh, we have minus two x squared. We have plus 16 X, and we have minus 32. So there is the answer that will have the smallest cleat coefficient here of one. And there's our polynomial.
Jesse T. The 1st problem is: 3, 2, -2 & the 2nd problem is: 3, 1, -2, -4 More
1 Expert Answer
If a polynomial has zeros at 3, 2 and -2 then this means that
(x-3), (x-2), and (x+2) are all factors of the polynomial
If you multiply these factors together you will get a polynomial with the given zeros
(x-3) * (x-2) * (x+2)
(x-3) * (x^2 - 4)
x^3 -3x^2 - 4x + 12
If you plug in 3, 2 or -2 for x you will see that the polynomial is equal to 0 at those points
The 2nd problem is similar but you will have 4 factors to multiply together
Still looking for help? Get the right answer, fast.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
If the zero is c, the factor is (x-c).
So for zeros of #-3,-1/3, 5#, the factors are
#(x+3)(x+1/3)(x-5)#
Let's take a look at the factor #(x+color(blue)1/color(red)3)#. Using the factor in this form will not result in integer coefficients because #1/3# is not an integer.
Move the #color(red)3# in front of the x and leave the #color(blue)1# in place: #(color(red)3x+color(blue)1)#.
When set equal to zero and solved, both
#(x+1/3)=0# and #(3x+1)=0# result in #x=-1/3#.
#f(x)=(x+3)(3x+1)(x-5)#
Multiply the first two factors.
#f(x)=(3x^2+10x+3)(x-5)#
Multiply/distribute again.
#f(x)=3x^3+10x^2+3x-15x^2-50x-15#
Combine like terms.
#f(x)=3x^3-5x^2-47x-15#