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taurus [48]
3 years ago
9

Which polynomial function has a leading coefficient of 1, roots -2 and 7 with multiplicity 1, and root 5 with multiplicity 2?

Mathematics
1 answer:
konstantin123 [22]3 years ago
4 0

Option D: f(x) = (x - 7)(x - 5)(x - 5)(x + 2) is the polynomial

Explanation:

Given that we need to determine the polynomial that has a leading coefficient of 1, roots -2 and 7 with multiplicity 1 and root 5 with multiplicity 2

Option A: f(x) = 2(x + 7)(x + 5)(x - 2)

The polynomial has roots -7, -5 and 2 with multiplicity 1.

Hence, Option A is not the correct answer.

Option B:  f(x) = 2(x - 7)(x - 5)(x + 2)

The polynomial has roots 7,5 and -2 with multiplicity 1.

Hence, Option B is not the correct answer.

Option C: f(x) = (x + 7)(x + 5)(x + 5)(x - 2)

The polynomial has roots 2 and -7 with multiplicity 1 and root -5 with multiplicity 2.

Hence, Option C is not the correct answer.

Option D: f(x) = (x - 7)(x - 5)(x - 5)(x + 2)

The polynomial has roots -2 and 7 with multiplicity 1 and root 5 with multiplicity 2.

Hence, Option D is the correct answer.

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Explanation:

Think of 25 as 25/1 and 5 as 5/1

Choice E is really saying \frac{25}{1} \div \frac{5}{1} which becomes \frac{25}{1} \times \frac{1}{5}. Note the second fraction flips and we change to a multiplication sign.

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Or you could think of it like this:

25 \times \frac{1}{5} = \frac{25}{1} \times \frac{1}{5} = \frac{25*1}{1*5} = \frac{25}{5} = 25 \div 5

to help see why the answer is E.

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