Question

Find an equation of a degree 3 polynomial (in factored form) with the given zeros of f(x): -5, -3,1.

Assume the leading coefficient is 1.

Answer 1

f(x) = x^3 + 7x^2 + 7x -15

Explanation:

f(x) = (x+5)(x+3)(x-1)

f(x) = (x^2 + 3x + 5x + 15)(x-1)

f(x) = (x^2 + 8x + 15)(x-1)

f(x) = x^3 + 8x^2 + 15x -x^2 -8x -15

simplify

f(x) = x^3 + 7x^2 + 7x -15

Answer 2

`f(x) = (x + 5)\, (x + 3)\, (x - 1)`.

Explanation:

By the factor theorem, if a constant
`a` is zero of the polynomial
`f(x)`,
`(x - a)` would be a factor of this polynomial. (Notice how
`x = a` would indeed set the value of
`(x - a)\!` to
`0`.)

For instance, since
`(-5)` is a zero of the polynomial
`f(x)`,
`(x - (-5))` would be a factor of
`f(x)\!`. Simplify this expression to get
`(x + 5)`.

Likewise, the zero
`(-3)` would correspond to the factor
`(x + 3)`, while the zero
`1` would correspond to the factor
`(x - 1)`.

All three of these factors above are linear, and the degree of the variable
`x` in each factor is
`1`. Multiplying three such linear factors would give a polynomial of degree
`3`.

Given the three factors, the expression of
`f(x)` in factored form would be:

`f(x) = m\, (x + 5)\, (x + 3)\, (x - 1)` for some constant
`m`.

When this expression is expanded, the constant
`m` would be the coefficient of the
`x^(3)` term (the leading term.) In other words,
`m\!` is the leading coefficient of
`f(x)`. This question has required this coefficient to be
`1`. Thus,
`m = 1`. The expression of
`f(x)\!` in factored form would be:

`f(x) = (x + 5)\, (x + 3)\, (x - 1)`.