Need with this math problem find a polynomial function of least degree having only real coefficients, a leading coefficient of 1, and roots of 1 minus startroot 3 endroot comma 1 plus startroot 3 endroot, and 7minus i.

P(x) = x^4 -16x^3 +76x^2 -72x -100

Step-by-step explanation:

The two roots 1-√3 and 1+√3 give rise to the quadratic factor ...

... (x -(1-√3))(x -(1+√3)) = (x-1)^2 -(√3)^2 = x^2 -2x -2

The complex root 7-i has a conjugate that is also a root. These two roots give rise to the quadratic factor ...

... (x -(7 -i))(x -(7 +i)) = (x-7)^2 -(i)^2 = x^2 -14x +50

The product of these two quadratic factors is ...

... P(x) = (x^2 -2x -2)(x^2 -14x +50) = x^4 +x^3(-14 -2) +x^2(50 +28 -2) +x(-100+28) -100

... P(x) = x^4 -16x^3 +76x^2 -72x -100

Please ignore - I missed the part about coefficients must be real and did not account for the conjugate root 7+i. Sorry!

Step-by-step explanation:

Three roots, 1-sqrt3, 1+sqrt3 and 7-i, are given so the polynomial is of third degree.

To have leading coefficient of 1, simply the term (x-root) together for given roots:

P(x) = (x-(1-sqrt3)) * ((x-(1+sqrt3)) * (x-(7-i))

= (x^2-(1-sqrt3+1+sqrt3)x+(1-sqrt3)(1+sqrt3)) * (x-7+i)

= (x^2-2x+1-3) * (x-7+i)

= (x^2-2x-2) * (x-7+i)

= x^3 - (9-i)x^2 + (12-2i)x + (14-2i)

if x is real, then it can be re-arranged to real and imaginary term:

P(x) = x^3 - 9x^2 + 12x + 14 + (x^2 - 2 x - 2)i

11 feet 6 inches and 8 feet 6 inches

50-38=12

step-by-step explanation: