Question

Find a degree 5 Polynomial p(x) with Zeros: 1,2,3,5+i and p(6)=240

Answer 1

need help with problem too?

Answer 2

If one is a zero, then
(x - 1) is a factor.

Answer 3

If those are the zeros, then the factors from which they came are (x - 1)(x - 2)(x - 3)(x - (5 + i))

Answer 4

The last one I a working on.

Answer 5

ohh okay so then it would be (x-1)(x-2)(x-3)(x-5-i)?

Answer 6

and i would use the synthetic division for P(6)

Answer 7

hmmm

Answer 8

P(6) means that when you put 6 in for x, your polynomial = 240. So the last factor I would say is (x - 6). Let's do that math now with all our factors. We will save the "i" one for last. Ok?

Answer 9

keep in mind that "complex roots" do not come all by themselves
they have a companion that comes with them, their "conjugate"

thus x = 5+i => x -5 -i=0

has its companion of \(\bf x-5+i\)

and thus would be the 5 roots

Answer 10

or I should put them as (x-5) -i and (x-5) + i

Answer 11

@IMStuck OKAY

Answer 12

@jdoe0001 SO then i would have those two as well

Answer 13

\(\bf [(x-5) -i][(x-5) + i]\qquad recall\implies {\color{brown}{ (a-b)(a+b) = a^2-b^2}}
\\ \quad \\\
[(x-5) -i][(x-5) + i]\implies [(x-5)^2-i^2]
\\ \quad \\
{\color{brown}{ i^2\to \sqrt{-1}\cdot \sqrt{-1}\to \sqrt{(-1)^2}\to -1}}
\\ \quad \\\
[(x-5)^2-i^2]\implies [(x-5)^2-({\color{brown}{ -1}})]\implies [(x-5)^2+1]\)

Answer 14

@jdoe0001 oh okay

Answer 15

and of course, to get the polynomial, as IMStuck said. just multiply them all

Answer 16

what do you mean multiply them all?

Answer 17

\(\bf (x-1)(x-2)(x-3)[(x-5)^2+1]=0
\\ \quad \\
(x-1)(x-2)(x-3)[(x-5)^2+1]=\textit{original polynomial}\)

Answer 18

okay then i would get :1,2,3,

Answer 19

and (x^2-10x+26)

Answer 20

yeap, and to get the original polynomial. you'd just multiply all roots so \(\bf (x-1)(x-2)(x-3)(x^2-10x+26)=\textit{original polynomial}\)

Answer 21

SO it's x^5-16x^4+97x^3-272x^2+346x-156

Answer 22

lemme check

Answer 23

okay

Answer 24

\(\bf {\color{brown}{ (x-1)(x-2)(x-3)}}(x^2-10x+26)
\\ \quad \\
{\color{brown}{ (x^3-6x^2+11x-6)}}(x^2-10x+26)\\
x^5-6x^4+11x^3-6x^2-10x^4+60x^3-110x^2+60x\lrcorner\\
+26x^3-156x^2+286-156
\\ \quad \\
{\color{blue}{ x^5-16x^4+97x^3-272x^2+130}}\)

Answer 25

hmmm wait.. a sec.. I'm mising the "x" bit =)

Answer 26

okay

Answer 27

:)

Answer 28

hmm anyow... yes,.. your correct :)

Answer 29

sorta ate the "x" for a sec =)

Answer 30

\(\bf {\color{brown}{ (x-1)(x-2)(x-3)}}(x^2-10x+26)
\\ \quad \\
{\color{brown}{ (x^3-6x^2+11x-6)}}(x^2-10x+26)\\
x^5-6x^4+11x^3-6x^2-10x^4+60x^3-110x^2+60x\rgroup\\
+26x^3-156x^2+286x-156
\\ \quad \\
{\color{blue}{ x^5-16x^4+97x^3-272x^2+346x-156}}\)

Answer 31

so what do i do with p(6)=240

Answer 32

hmm

Answer 33

@IMStuck Hey do you know what do i do with p(6)=240

Answer 34

so I just ran it through, setting x = 6 and gave me p(x) or p(6) \(\bf p(x)={\color{blue}{ x^5-16x^4+97x^3-272x^2+346x-156}}
\\ \quad \\
p(6)=120\qquad thus
\\ \quad \\
p(x)=2({\color{blue}{ x^5-16x^4+97x^3-272x^2+346x-156}})
\\ \quad \\
p(6)=2(120)\to 240\)

Answer 35

ahemm it gave 120 obviously =)

Answer 36

so that means the polynomial originally has a multiplier of 2 in front

Answer 37

so you just found a number that gave you 240?

Answer 38

why did you use those two number why not 3(80)

Answer 39

ohh because

if you plug "6" as the "input" for "x" that is, set x = 6

the polynomial of \(\bf x^5-16x^4+97x^3-272x^2+346x-156\) gives 120

but we know that p(6) is twice that much

that means, that the polynomial is being multiplied by a digit of 2

Answer 40

ohh okay i get it now :)

Answer 41

Thanks!!

Answer 42

yw