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3 years ago

Which second degree polynomial function f(x) has a lead coefficient of 3 and roots 4 and 1?

Mathematics

2 answers:

Stella [2.4K]3 years ago

7 0

Answer:

f(x) = 3x² - 15x + 12

Step-by-step explanation:

Given f(x) has roots x = a and x = b, then

(x - a) and (x - b) are the factors

f(x) is then the product of the factors

f(x) = a(x - a)(x - b) ← where a is a multiplier

Given roots are x = 4 and x = 1, then

(x - 4) and (x - 1) are the factors

With a = 3, then

f(x) = 3(x - 4)(x - 1) ← expand factors using FOIL

= 3(x² - 5x + 4) ← distribute

= 3x² - 15x + 12

Sophie [7]3 years ago

5 0

Answer:

f(x) = 3x^2 - 15x + 12

Step-by-step explanation:

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