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Which second-degree polynomial function f (x) has a lead coefficient of 4 and roots 5 and 2?

2 Answers

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Answer:

d

Explanation:

User Vvo
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6 votes

Answer:

The second degree polynomial is f(x) = 4x² - 28x + 40

Explanation:

* Lets revise the general form of the second-degree polynomial

- The general form of the second degree polynomial is

f(x) = ax² + bx + c, where a , b , c are constant

- The highest power of the variable that occurs in the polynomial

is called the degree of a polynomial.

- The leading term is the term with the highest power, and its

coefficient is called the leading coefficient.

- The leading coefficient is the coefficient of x²

∴ a = 4

∴ f(x) = 4x² + bx + c

- The roots of a polynomial are also called its zeroes, because

the roots are the x values at which the function equals zero

∴ When f(x) = 0, the values of x are 5 and 2

* To find the value of b and c substitute the values of x in f(x) = 0

- At x = 5

∵ 4(5)² + b(5) + c = 0 ⇒ simplify it

∴ 100 + 5b + c = 0 ⇒ subtract 100 from both sides

∴ 5b + c = -100 ⇒ (1)

- At x = 2

∵ 4(2)² + b(2) + c = 0 ⇒ simplify it

∴ 16 + 2b + c = 0 ⇒ subtract 16 from both sides

∴ 2b + c = -16 ⇒ (2)

- Subtract (2) from (1)

∴ 3b = -84 ⇒ divide both sides by 3

∴ b = -28

- Substitute the value of b in (1) or (2) to find c

∵ 2(-28) + c = -16

∴ -56 + c = -16 ⇒ add 56 to both sides

∴ c = 40

∴ f(x) = 4x² - 28x + 40

* The second degree polynomial is f(x) = 4x² - 28x + 40

User Ivan Mir
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