Question

MEDAL AND FAN

Answer 1

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. (1 point)

2, -4, and 1 + 3i

f(x) = x4 - 2x2 + 36x - 80
f(x) = x4 - 3x3 + 6x2 - 18x + 80
f(x) = x4 - 9x2 + 36x - 80
f(x) = x4 - 3x3 - 6x2 + 18x - 80
18. Using the given zero, find all other zeros of f(x). (1 point)

-2i is a zero of f(x) = x4 - 45x2 - 196

2i, 14i, -14i
2i, 7, -7
2i, 14, -14
2i, 7i, -7i

Answer 2

A zero of a polynomial is an x-value that results in a value of zero fir the function, i.e. an x-intercept. Perhaps the easiest way to tackle this problem is to try each given x-value and select the ones that result in f(x) = 0

Answer 3

I have no idea

Answer 4

@ospreytriple

Answer 5

Take the first option in no. 17. Try x=2\[f(x)=x^4-2x^2+36x-80 = 2^4-2(2)^2+36(2)-80 = 0\]So x=2 is a zero. Now you have to check the rest.

Answer 6

a polynomial with roots r_1, r_2, r_3... r_n can be written as
f(x) = ( x - r_1)*( x - r_2)*( x - r_3 ) ... ( x - r_n)

Answer 7

no this wouldn't be the case. @chappy0729 may i start the qyuestion?

Answer 8

here in this case complex number is also there so for determination of this degree will be on the basis of this.

Answer 9

You are given roots : 2, -4, and 1 + 3i
Note that complex roots come in conjugate pairs. so your roots are actually
2, -4, ,1 + 3i, 1 - 3i

f(x) = ( x - 2 ) * ( x - (-4) ) * ( x - (1 + 3i ) ) * ( x - ( 1 - 3i ) )

now you have to expand that

Answer 10

when a+ib is one root then there must be other root like a-ib complex numbers exist in conjugate roots.

Answer 11

the polynomial would be :
(x-2)(x+4)(x-1+3i)(x-1-3i)

Answer 12

@chappy0729 did u get it?

Answer 13

yes thank you :) @ribhu

Answer 14

one zero is given as 2i so other becomes -2i
and this polynomial becomes (x^2+4)

Answer 15

then divide the 4 degree polynomial with this (x^2+4) and get another polynomial. factorise it and then obtain the remaining root of the equation.
just remember complex roots always exist in conjugates.

Answer 16

@chappy0729 i hope i made myself very clear.

Answer 17

@chappy0729 did u get it?