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Which second degree polynomial function f(x) has a lead coefficient of 3 and roots 4 and 1? f(x) = 3x2 + 5x + 4 f(x) = 3x2 + 15x + 12 f(x) = 3x2 – 5x + 4 f(x) = 3x2 – 15x + 12

User Dwilkins
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2 Answers

3 votes

Answer:

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

Explanation:

its Britany bitc

Explanation:

User PlunkettBoy
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7 votes

Answer:

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

Explanation:

All of the second degree polynomials listed have a leading coefficient of 3.

Since we are asked which has roots 4 and 1, we know we need to find the polynomial where x is a negative number, then solve for x.

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

= 3(x^2 - 5x + 4) --- Here, I factored out the 3.

= 3((x-4)(x-1)) --- solve for the quadratic.

Therefore, solving for x, we have

x - 4 = 0 and x - 1 = 0

x = 4 x = 1

Thus, this polynomial has a lead coefficient of 3 and its roots are 4 and 1.

User Pouzzler
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