Find a polynomial function of least degree having only real coefficients, a leading of 1, and zeros of 6 and 3 + i.

Since we know that complex solutions ALWAYS come in pairs, the minimal polynomial must include the root of 4-i as an acceptable root..

This leads to a polynomial of ...

p(x) = (x - 5)(x - (4+i))(x - (4-i))

Now we can simplify...

(x - (4+i))(x - (4-i)) --> x2 + [(4+i)+(4-i)]x + (4+i)(4-i) --> x2 + 8x + [ 16 -4i + 4i -i2 ] = x2 + 8x +17

Take this quadratic and multiply by (x-5) to get the final, minimal polynomial...

p(x) = (x-5)(x2 + 8x +17) = x3 + 8x2 + 17x -5x2 + 40x - 85 = x3 + 3x2 + 57x - 85

So p(x) = x3 + 3x2 + 57x - 85

p(x) has only real coefficients, the leading coefficient of 1, and is the smallest to allow for the two solutions requested

x+14

step-by-step explanation:

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|