Use irrational conjugate theorem to write a polynomial function of least degree that has rational coefficients, a leading coefficient of 1 and given zeros.

Hhh

Step-by-step explanation:

f(x) = x³-11x²+32x-24

Step-by-step explanation:

zeros = 3 and 4+√8

because of the irrational conjugate theorem it = 3, 4+√8, 4-√8

f(x) = (x-3) (x-(4+√8)) (x-(4-√8))

f(x) = (x-3) ((x-4)-√8) ((x-4)+√8)

                             (a-b) (a+b)

                             = a²-b²

f(x) = (x-3) ((x-4)² - (√8)²)

f(x) = (x-3) ((x-4)² - 8)

f(x) = (x-3) ((x²-8x+16)-8)

f(x) = x³-11x²+32x-24

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zeros=3, 4+√8→ 3, 4+√8, 4-√8

f(x)=(x-3)(x-(4+√8))(x-(4-√8)

f(x)=(x-3)((x-4)-√8)((x-4)+√8)

f(x)=(x-3)((x-4)2-(√8)2)

f(x)=(x-3)((x2-8x+16)-8)

f(x)=x3-112+32x-24

Step-by-step explanation:

slope = Negative one-fourth, y-intercept = 1

Step-by-step explanation: