Question

Form a polynomial whose real zeros and degrees are given

Zeros: -2, 0, 7
Degree: 3
Type a polynomial with integer coefficients and a leading coefficient of 1
f(x)= ?

Answer 1

Given ,

`\bold\red{zeroes \: of \: a \: cubic \: polynomial \: are \: - 2 \:, \: 0 \: and \: 7} \\`

`let \:, \\ \boxed{\alpha \: = - 2} \\ \boxed{\beta \: = 0} \\ \boxed{\gamma \: = 7}`

Now ,

`sum \: of \: zeroes \: = \alpha \: + \: \beta \: + \: \gamma \: \\ \dashrightarrow \: sum = - 2 + 0 + 7 = \underline{5} \\ \\ sum \: of \: product \: of \: zeroes \: taken \: two \: at \: a \: time \: = \alpha\beta \: + \beta\gamma \: + \: \alpha\gamma \\ \dashrightarrow \: ( - 2)(0) \: + \: (0)(7) \: + \: (7)( - 2) \: = \: \underline{ - 14} \\ \\ product \: of \: zeroes \: = \: \alpha\beta\gamma \\ \dashrightarrow \: ( - 2)(0)(7) = \underline{0}`

We know that ,

`\boxed{cubic \: polynomial \: = {x}^(3) - (\alpha \: + \: \beta \: + \: \gamma \:) \:{x}^(2) + (\alpha\beta \: + \: \beta\gamma \: + \: \alpha\gamma) \:x \: - \: \alpha\beta\gamma}`

Plugging in the values , we get

`\boxed{f(x) = {x}^(3) - 5 {x}^(2) - 14x + 0}`

since this polynomial has degree 3 , it is called a cubic polynomial.

hope helpful! :)