Answered

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What is the leading coefficient of a third degree function that has an output of 221 when x=2, and has zeros of −15, 3i, and −3i?

Sagot :

sqdancefan

Answer:

  1

Step-by-step explanation:

The function with the given zeros will factor as ...

  f(x) = a(x +15)(x^2 +9) . . . . with leading  coefficient 'a'

You have ...

  f(2) = 221 = a(2+15)(2^2+9) = a(17)(13) = 221a

Then a = 221/221 = 1

The leading coefficient is 1.

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Additional comment

As you know, a function with zero x=p has a factor of (x -p). The given zeros mean the function has factors (x -(-15)), (x -3i). and (x -(-3i)). The product of the last two factors is the difference of squares: (x^2 -(3i)^2) = (x^2 -(-9)) = (x^2 +9). This is how we arrived at the factorization shown above.

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