Math, asked by martinvega, 1 year ago

Which polynomial function has a leading coefficient of 1, roots –2 and 7 with multiplicity 1, and root 5 with multiplicity 2?
a) f(x) = 2(x + 7)(x + 5)(x – 2)
b) f(x) = 2(x – 7)(x – 5)(x + 2)
c) f(x) = (x + 7)(x + 5)(x + 5)(x – 2)
d) f(x) = (x – 7)(x – 5)(x – 5)(x + 2)

Answers

Answered by TooFree
9

Option a:

f(x) = 2(x + 7)(x + 5)(x – 2)

⇒ The leading coefficient is 2

⇒ Not the answer we are looking for


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Option b:

f(x) = 2(x – 7)(x – 5)(x + 2)

⇒ The leading coefficient is 2

⇒ Not the answer we are looking for


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Option c:

f(x) = (x + 7)(x + 5)(x + 5)(x – 2)

⇒ Leading coefficient = 1


Find the roots:

(x + 7)(x + 5)(x + 5)(x – 2) = 0

(x + 7)(x + 5)²(x – 2) = 0

x = -7 or x = -5 or x = 2

⇒ The roots are 2  and -7 with multiplicity 1, and root - 5 with multiplicity 2

⇒ Not the answer we are looking for


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Option d:

f(x) = (x – 7)(x – 5)(x – 5)(x + 2)

⇒ Leading coefficient = 1


Find the roots:

(x – 7)(x – 5)(x – 5)(x + 2) = 0

(x – 7)(x – 5)²(x + 2) = 0

x = 7 or x = 5 or x = -2

⇒ The roots are -2  and 7 with multiplicity 1, and root 5 with multiplicity 2

⇒ This is the answer we are looking for


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Answer: (d) f(x) = (x – 7)(x – 5)(x – 5)(x + 2)

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