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Wewaii [24]
3 years ago
13

Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms. Third-degree, with zeros of −3

, −1, and 2, and passes through the point (1,10).
Mathematics
1 answer:
RSB [31]3 years ago
8 0

Answer:

\Large \boxed{-\dfrac{5}{4}(x+3)(x+1)(x-2)}

Step-by-step explanation:

Hello,

Based on the indication, we can write this polynomial as below, k being a real number that we will have to identify (degree = 3 and we have three zeroes -3, -1, and 2).

   \Large \boxed{k(x+3)(x+1)(x-2)}

We know that the point (1,10) is on the graph of this function, so we can say.

k(1+3)(1+1)(1-2)=10}\\\\4*2*(-1)*k=10\\\\-8k=10\\\\k=\dfrac{10}{-8}=-\dfrac{5}{4}

Then the solution is:

\large \boxed{-\dfrac{5}{4}(x+3)(x+1)(x-2)}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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<h2>Answer and Explanation to questions 16,17,18</h2>

∠3 is supplementary to ∠1 means: ∠3 + ∠1 = 180°

And, according to figure ∠1 + ∠2 = 180° as ∠1 and ∠2 form a straight line.

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<h2>Answer and Explanation to questions 19</h2>

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