The function f(x) is a quartic function and the zeros of f(x) are -4,-3,2 and 5. Assume the leading coefficient of f(x) is 1. Write the equation of the quartic polynomial in standard form

The equation of the quartic polynomial in standard form is,

⇒ f (x) = x⁴ - 27x² - 14x + 120

An algebraic equation with the second degree of the variable is called an Quadratic equation.

We have to given that;'

The function f(x) is a quartic function and the zeros of f(x) are -4,-3,2 and 5.

And, the leading coefficient of f(x) is 1.

Hence, We get;

The equation of the quartic polynomial in standard form is,

⇒ f (x) = 1 (x - (- 4)) (x - (- 3)) (x - 2) (x - 5)

⇒ f (x) = (x + 4) (x + 3) (x - 2) (x - 5)

⇒ f (x) = (x² + 7x + 12) (x² - 7x + 10)

⇒ f (x) = x⁴ - 7x³ + 10x² + 7x³ - 49x² + 70x + 12x² - 84x + 120

⇒ f (x) = x⁴ - 27x² - 14x + 120

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