You can
put this solution on YOUR website!(x - 1.2)(x - 4)
x^2 - 1.2x - 4x + 4.8
x^2 - 5.2x + 4.8 = f(x) i do not know where your last zero is, so i will just do one
(x^2 - 5.2x + 4.8)(x + 5)
x^3 - 5.2x^2 + 4.8x + 5x^2 - 26x + 24
x^3 - 0.2x^2 - 21.2x + 24 = f(x)
we need the last digit to be sixteen:
24x = -16
x = -16/24 = -8/12 = -4/6 = -2/3
(x^3 - 0.2x^2 - 21.2x + 24)(-2/3) = f(x)
(-2/3)x^3 + (2/15)x^2 + (212/15)x - 16 = f(x)
You can
put this solution on YOUR website!Find the polynomial f(x) of degree
three that has zeroes at 1.2 and 4 such that f(0) = -16.
Unlike the previous tutor, I'm assuming " 1.2 " was a typo,
and you really meant the "." to be a comma rather than a decimal.
This, in other words:
Find the polynomial f(x) of degree three that has zeroes
at 1, 2 and 4 such that f(0) = -16.
All polynomials of degree n that have zeros r1, r2, ..., rn
are of the form
f(x) = k(x-r1)(x-r2)···(x-rn)
where k can be any non-zero number.
So any polynomial of degree 3 that has zeros 1, 2, and 4 is
of the form
f(x) = k(x-1)(x-2)(x-4)
where k can be any non-zero number.
But we also want f(0) to equal -16. So it has to be true that
if we substitute 0 for x and then -16 for f(0), they should be
equal. So,
f(x) = k(x-1)(x-2)(x-4)
f(0) = k(0-1)(0-2)(0-4)
-16 = -8k
2 = k
So substitute 2 for k in
f(x) = k(x-1)(x-2)(x-4)
to give
f(x) = 2(x-1)(x-2)(x-4)
Multiply all that out and you'll get
f(x) = 2x³ - 14x² + 28x - 16
Edwin
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